GSEB Class 10 Maths Solutions Chapter 6 Triangle Exercise 6.3

Question 1. State in which pairs of triangles in the figures are similar. Write the similarity criteria used by you for answering the question and also write the pairs of similar triangles in the symbolic form.
Answer:
(i) In \( \triangle ABC \) and \( \triangle PQR \):
\( \angle A = \angle P \) (each \( 60^\circ \))
\( \angle B = \angle Q \) (each \( 80^\circ \))
\( \angle C = \angle R \) (each \( 40^\circ \))
The corresponding angles of \( \triangle ABC \) are equal to the corresponding angles of \( \triangle PQR \).
\( \implies \triangle ABC \sim \triangle PQR \) (by AAA Similarity)
(ii) In \( \triangle ABC \) and \( \triangle QRP \):
\( \frac{AB}{QR} = \frac{2}{4} = \frac{1}{2} \)
\( \frac{BC}{RP} = \frac{2.5}{5} = \frac{1}{2} \)
\( \frac{CA}{PQ} = \frac{3}{6} = \frac{1}{2} \)
Since the ratios of corresponding sides are equal, \( \frac{AB}{QR} = \frac{BC}{RP} = \frac{CA}{PQ} \).
\( \implies \triangle ABC \sim \triangle QRP \) (by SSS Similarity)
(iii) In \( \triangle LMP \) and \( \triangle DEF \):
\( \frac{LM}{DE} = \frac{2.7}{4} \)
\( \frac{MP}{EF} = \frac{2}{5} \)
\( \frac{LP}{DF} = \frac{3}{6} = \frac{1}{2} \)
Here, \( \frac{LM}{DE} \ne \frac{MP}{EF} \ne \frac{LP}{DF} \).
\( \implies \triangle LMP \) and \( \triangle DEF \) are not similar.
(iv) In \( \triangle MNL \) and \( \triangle PQR \):
\( \frac{ML}{QR} = \frac{5}{10} = \frac{1}{2} \)
\( \frac{MN}{QP} = \frac{2.5}{5} = \frac{1}{2} \)
\( \angle NML = \angle PQR \) (each \( 70^\circ \))
\( \implies \triangle MNL \sim \triangle QPR \) (by SAS Similarity)
(v) In \( \triangle ABC \) and \( \triangle DEF \):
\( \angle A = \angle F \) (each \( 80^\circ \))
The included sides of \( \angle A \) in \( \triangle ABC \) are not provided, while the included sides for \( \angle F \) in \( \triangle DEF \) are given.
Therefore, the SAS similarity property will not be applicable in this specific case, and no other similarity criterion is met from the given information.
(vi) In \( \triangle DEF \) and \( \triangle PQR \):
For \( \triangle DEF \):
\( \angle F = 180^\circ - (70^\circ + 80^\circ) = 180^\circ - 150^\circ = 30^\circ \)
For \( \triangle PQR \):
\( \angle P = 180^\circ - (80^\circ + 30^\circ) = 180^\circ - 110^\circ = 70^\circ \)
Now, comparing \( \triangle DEF \) and \( \triangle PQR \):
\( \angle D = \angle P \) (each \( 70^\circ \))
\( \angle E = \angle Q \) (each \( 80^\circ \))
\( \angle F = \angle R \) (each \( 30^\circ \))
Since all corresponding angles are equal,
\( \implies \triangle DEF \sim \triangle PQR \) (by AAA Similarity)
In simple words: We check if triangles are similar by looking at their angles or the ratios of their sides. If all angles are the same (AAA), or all side ratios are the same (SSS), or two sides have the same ratio and the angle between them is the same (SAS), then the triangles are similar. We write similar triangles with the matching vertices in order.

Exam Tip: When proving triangle similarity, clearly state the corresponding angles or sides and the similarity criterion (AAA, SSS, or SAS) used. Ensure the vertices are listed in the correct corresponding order.

Question 2. In the figure \( \triangle ODC \sim \triangle OBA \), \( \angle BOC = 125^\circ \) and \( \angle CDO = 70^\circ \). Find \( \angle DOC \), \( \angle DCO \) and \( \angle OAB \).
Answer: We have the following information from the problem:
\( \triangle ODC \sim \triangle OBA \)
\( \angle BOC = 125^\circ \)
\( \angle CDO = 70^\circ \)
First, we find \( \angle DOC \). Since \( \angle DOC \) and \( \angle BOC \) form a linear pair (angles on a straight line):
\( \angle DOC + \angle BOC = 180^\circ \)
\( \angle DOC + 125^\circ = 180^\circ \)
\( \angle DOC = 180^\circ - 125^\circ \)
\( \angle DOC = 55^\circ \)........(1)
Next, we find \( \angle DCO \). In \( \triangle DOC \), the sum of angles is \( 180^\circ \) (angle sum property of a triangle):
\( \angle DCO + \angle CDO + \angle DOC = 180^\circ \)
\( \angle DCO + 70^\circ + 55^\circ = 180^\circ \)
\( \angle DCO + 125^\circ = 180^\circ \)
\( \angle DCO = 180^\circ - 125^\circ \)
\( \angle DCO = 55^\circ \)........(2)
Finally, we find \( \angle OAB \). Since it is given that \( \triangle ODC \sim \triangle OBA \), their corresponding angles must be equal:
\( \angle OCD = \angle OAB \) (corresponding angles of similar triangles)
From equation (2), we know \( \angle DCO = 55^\circ \). Therefore,
\( \angle OAB = 55^\circ \)........(3)
Hence, the required angles are:
\( \angle DOC = 55^\circ \)
\( \angle DCO = 55^\circ \)
\( \angle OAB = 55^\circ \)
In simple words: We used two basic geometry rules to find the missing angles. First, angles on a straight line add up to 180 degrees. Second, angles inside a triangle add up to 180 degrees. Because the two triangles are similar, their matching angles are the same.

Exam Tip: Remember to clearly state the geometric properties or theorems (e.g., linear pair, angle sum property, corresponding angles of similar triangles) you use at each step of your solution to ensure full marks.

Question 3. Diagonal AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \( \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}} \).
Answer: We have a trapezium ABCD where AB is parallel to DC. The diagonals AC and BD meet at point O.
Now, consider \( \triangle OAB \) and \( \triangle OCD \).
Since AB || DC (given):
\( \angle OBA = \angle ODC \) (These are alternate interior angles)
\( \angle OAB = \angle OCD \) (These are also alternate interior angles)
\( \angle AOB = \angle COD \) (These are vertically opposite angles)
By the Angle-Angle (AA) similarity criterion, if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
Therefore, \( \triangle OAB \sim \triangle OCD \) (by AA similarity)
When two triangles are similar, the ratios of their corresponding sides are equal.
\( \implies \frac{OA}{OC} = \frac{OB}{OD} = \frac{AB}{CD} \)
Hence, we have shown that \( \frac{OA}{OC} = \frac{OB}{OD} \).
In simple words: We looked at two small triangles inside the trapezium, made by the diagonals crossing. Since the parallel sides of the trapezium make special "alternate interior angles" that are equal, and the angles where the diagonals cross are also equal (vertically opposite), these two small triangles are similar. When triangles are similar, their sides are in proportion, which proved the relationship.

Exam Tip: For problems involving parallel lines, always look for alternate interior angles, corresponding angles, or vertically opposite angles to establish angle equality, which is often the first step in proving triangle similarity.

Question 4. In the given figure, \( \frac {QR}{QS} = \frac {QT}{PR} \) and \( \angle 1 = \angle 2 \), show that \( \triangle PQS \sim \triangle TQR \).
Answer: We are given the following:
1. \( \frac {QR}{QS} = \frac {QT}{PR} \)
2. \( \angle 1 = \angle 2 \)
Consider \( \triangle PQR \).
Since \( \angle 1 = \angle 2 \), the sides opposite to these equal angles must also be equal.
Therefore, \( PR = PQ \) (Sides opposite to equal angles in a triangle are equal).....(1)
Now, substitute \( PQ \) for \( PR \) in the given ratio:
\( \frac {QR}{QS} = \frac {QT}{PR} \)
Using (1), this becomes:
\( \frac {QR}{QS} = \frac {QT}{PQ} \).....(2)
Now, let's consider \( \triangle PQS \) and \( \triangle TQR \).
From equation (2), we have the ratio of corresponding sides:
\( \frac {QR}{QS} = \frac {QT}{PQ} \)
Also, both triangles share a common angle, \( \angle Q \). So,
\( \angle PQS = \angle TQR \) (This is the common angle, \( \angle Q \), which is \( \angle 1 \)).
By the Side-Angle-Side (SAS) similarity criterion, if two sides of one triangle are proportional to two sides of another triangle and the included angles are equal, then the two triangles are similar.
Thus, \( \triangle PQS \sim \triangle TQR \) (by SAS similarity).
In simple words: We used the given angle information to show that two sides of one triangle were actually equal. Then we used this fact to adjust the given side ratio. With this new ratio and the common angle they shared, we could prove the two triangles were similar using the SAS (Side-Angle-Side) rule.

Exam Tip: When given an isosceles triangle or equal angles, always deduce that the sides opposite those angles are equal. This substitution can be crucial for simplifying given ratios and applying similarity criteria.

Question 5. S and T are points on sides PR and QR of \( \triangle PQR \) such that \( \angle P = \angle RTS \). Show that \( \triangle RPQ \sim \triangle RTS \). (CBSE 2012)
Answer: We are given that T is a point on QR and S is a point on PR. Also, we know that \( \angle P = \angle RTS \).
Now, let's consider \( \triangle RPQ \) and \( \triangle RTS \).
We are given:
\( \angle RPQ = \angle RTS \) (These are the given equal angles).
Both triangles share a common angle:
\( \angle PRQ = \angle TRS \) (This is the common angle \( R \)).
By the Angle-Angle (AA) similarity criterion, if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
Therefore, \( \triangle RPQ \sim \triangle RTS \) (by AA similarity).
In simple words: We looked at the two triangles and saw that one angle from each was given as equal. They also shared one angle in common. Since two angles in both triangles were the same, we could say they were similar using the Angle-Angle rule.

Exam Tip: In similarity problems, always check for common angles that both triangles share. This often provides one of the necessary angle equalities for AA similarity.

Question 6. In figure, if \( \triangle ABE \cong \triangle ACD \), show that \( \triangle ADE \sim \triangle ABC \).
Answer: We are given that \( \triangle ABE \cong \triangle ACD \) (These are congruent triangles).
When two triangles are congruent, their corresponding parts are equal (CPCT).
Therefore, from \( \triangle ABE \cong \triangle ACD \):
\( AB = AC \) ........(1) (by CPCT)
\( AE = AD \) ........(2) (by CPCT)
Now, let's divide equation (1) by equation (2):
\( \frac{AB}{AE} = \frac{AC}{AD} \)
Rearranging this, we get:
\( \frac{AB}{AC} = \frac{AE}{AD} \)
Now, consider \( \triangle ADE \) and \( \triangle ABC \).
We have shown that the ratio of two sides is equal:
\( \frac{AD}{AB} = \frac{AE}{AC} \)
Both triangles share a common angle:
\( \angle DAE = \angle BAC \) (This is the common angle \( A \)).
By the Side-Angle-Side (SAS) similarity criterion, if two sides of one triangle are proportional to two sides of another triangle and the included angles are equal, then the two triangles are similar.
Therefore, \( \triangle ADE \sim \triangle ABC \) (by SAS similarity).
In simple words: We started by using the fact that the first two triangles were identical (congruent) to find equal sides. Then we divided these equal sides to find a constant ratio. Finally, using this side ratio and the common angle shared by \( \triangle ADE \) and \( \triangle ABC \), we proved their similarity using the SAS rule.

Exam Tip: When congruence is given, immediately use CPCT to establish equal sides or angles. These equalities are often key in proving subsequent similarity relationships.

Question 7. In the figure, altitude AD and CE of \( \triangle ABC \) intersect each other at P. Show that:
(i) \( \triangle AEP \sim \triangle CDP \)
(ii) \( \triangle ABD \sim \triangle CBE \)
(iii) \( \triangle AEP \sim \triangle ADB \)
(iv) \( \triangle PDC \sim \triangle BEC \)
Answer: We have \( \triangle ABC \) where AD and CE are altitudes that meet at point P. An altitude forms a \( 90^\circ \) angle with the base.
(i) In \( \triangle AEP \) and \( \triangle CDP \):
\( \angle AEP = \angle CDP \) (Both are \( 90^\circ \) as AD and CE are altitudes)
\( \angle APE = \angle CPD \) (These are vertically opposite angles)
\( \implies \triangle AEP \sim \triangle CDP \) (by AA Similarity)
(ii) In \( \triangle ABD \) and \( \triangle CBE \):
\( \angle ADB = \angle CEB \) (Both are \( 90^\circ \) as AD and CE are altitudes)
\( \angle ABD = \angle CBE \) (This is the common angle \( B \))
\( \implies \triangle ABD \sim \triangle CBE \) (by AA similarity)
(iii) In \( \triangle AEP \) and \( \triangle ADB \):
\( \angle AEP = \angle ADB \) (Both are \( 90^\circ \) as CE and AD are altitudes)
\( \angle PAE = \angle DAB \) (This is the common angle \( A \))
\( \implies \triangle AEP \sim \triangle ADB \) (by AA similarity)
(iv) In \( \triangle PDC \) and \( \triangle BEC \):
\( \angle PDC = \angle BEC \) (Both are \( 90^\circ \) as AD and CE are altitudes)
\( \angle PCD = \angle BCE \) (This is the common angle \( C \))
\( \implies \triangle PDC \sim \triangle BEC \) (by AA similarity)
In simple words: For each pair of triangles, we looked for two pairs of equal angles. Since AD and CE are altitudes, they create 90-degree angles. We also used vertically opposite angles and common angles. Once we found two equal angles in each pair, we used the Angle-Angle (AA) similarity rule to prove they were similar.

Exam Tip: When dealing with altitudes, remember that they always create a \( 90^\circ \) angle with the opposite side. This provides one angle for AA similarity. Then, look for common angles or vertically opposite angles.

Question 8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that \( \triangle ABE \sim \triangle CFB \).
Answer: We have a parallelogram ABCD. Side AD is extended to point E, and line segment BE cuts CD at point F.
Now, let's consider \( \triangle ABE \) and \( \triangle CFB \).
Since ABCD is a parallelogram, AB is parallel to DC (and thus to FC). Also, AD is parallel to BC (and thus AE is parallel to BC).
1. \( \angle BAE = \angle FCB \)
Since AB || DC, and AD is produced to E, we have AB || DF. Also, AE || BC.
\( \angle BAE \) (which is \( \angle DAB \)) and \( \angle FCB \) (which is \( \angle BCD \)) are opposite angles of a parallelogram, which are equal. So, \( \angle BAE = \angle BCD \). Given that F is on CD, \( \angle BCD \) is the same as \( \angle FCB \). Therefore, \( \angle BAE = \angle FCB \).
2. \( \angle AEB = \angle CBF \)
Since AE || BC (opposite sides of parallelogram ABCD, with AD extended to E), and BE is a transversal line.
These are alternate interior angles. Thus, \( \angle AEB = \angle CBF \).
By the Angle-Angle (AA) similarity criterion, if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
Therefore, \( \triangle ABE \sim \triangle CFB \) (by AA similarity).
In simple words: We used properties of parallelograms and parallel lines. Opposite angles in a parallelogram are equal. Also, when two parallel lines are crossed by another line, the alternate interior angles are equal. With two pairs of equal angles, we proved the two triangles were similar using the AA rule.

Exam Tip: In problems involving parallelograms, always use the properties of parallel lines (alternate interior angles, corresponding angles) and parallelogram properties (opposite angles are equal, opposite sides are parallel) to identify equal angles for similarity proofs.

Question 9. In the figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
(i) \( \triangle ABC \sim \triangle AMP \)
(ii) \( \frac {CA}{PA} = \frac {BC}{MP} \)
Answer: We are given two right-angled triangles, \( \triangle ABC \) and \( \triangle AMP \).
\( \triangle ABC \) is right-angled at B, so \( \angle ABC = 90^\circ \).
\( \triangle AMP \) is right-angled at M, so \( \angle AMP = 90^\circ \).
(i) To prove \( \triangle ABC \sim \triangle AMP \):
Consider \( \triangle ABC \) and \( \triangle AMP \).
1. \( \angle ABC = \angle AMP \) (Both angles are \( 90^\circ \)).
2. \( \angle BAC = \angle MAP \) (This is the common angle \( A \) shared by both triangles).
By the Angle-Angle (AA) similarity criterion, if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
Therefore, \( \triangle ABC \sim \triangle AMP \) (by AA similarity).
(ii) To prove \( \frac {CA}{PA} = \frac {BC}{MP} \):
Since we have already proven that \( \triangle ABC \sim \triangle AMP \) in part (i), the ratios of their corresponding sides must be equal.
The corresponding sides are:
\( \frac{AB}{AM} = \frac{BC}{MP} = \frac{AC}{AP} \)
From these ratios, we can extract the required relation:
\( \frac{BC}{MP} = \frac{AC}{AP} \)
Rearranging this slightly, we get:
\( \frac{CA}{PA} = \frac{BC}{MP} \)
Hence, proven.
In simple words: First, we used the fact that both triangles had a right angle and shared a common angle to prove they were similar. Second, because they were similar, we knew that their matching sides must be in proportion, which allowed us to show the given side ratio was correct.

Exam Tip: Always remember that corresponding parts of similar triangles are proportional. Once similarity is proven, immediately write down all possible ratios of corresponding sides. This makes it easy to pick out the specific ratio required by the question.

Question 10. CD and GH are respectively the bisectors of \( \angle ACB \) and \( \angle EGF \) such that D and H lie on sides AB and FE of \( \triangle ABC \) and \( \triangle FEG \) respectively. If \( \triangle ABC \sim \triangle FEG \) show that:
(i) \( \frac {CD}{GH} = \frac {AC}{FG} \)
(ii) \( \triangle DCB \sim \triangle HGE \)
(iii) \( \triangle DCA \sim \triangle HGF \)
Answer: We are given two similar triangles, \( \triangle ABC \) and \( \triangle FEG \). CD is the bisector of \( \angle ACB \), and GH is the bisector of \( \angle EGF \).
Since \( \triangle ABC \sim \triangle FEG \), their corresponding angles are equal, and their corresponding sides are proportional:
\( \angle A = \angle F \)
\( \angle B = \angle E \)
\( \angle C = \angle G \) (i.e., \( \angle ACB = \angle FGE \))
\( \frac{AB}{FE} = \frac{BC}{EG} = \frac{AC}{FG} \)
(i) To show \( \frac {CD}{GH} = \frac {AC}{FG} \):
Consider \( \triangle ACD \) and \( \triangle FGH \).
1. \( \angle A = \angle F \) (Corresponding angles of similar triangles \( \triangle ABC \sim \triangle FEG \)).
2. \( \angle ACB = \angle FGE \) (Corresponding angles of similar triangles \( \triangle ABC \sim \triangle FEG \)).
Since CD bisects \( \angle ACB \) and GH bisects \( \angle FGE \):
\( \frac{1}{2}\angle ACB = \frac{1}{2}\angle FGE \)
\( \implies \angle ACD = \angle FGH \)
By the Angle-Angle (AA) similarity criterion,
\( \implies \triangle ACD \sim \triangle FGH \) (by AA similarity).
Since these triangles are similar, the ratios of their corresponding sides are equal:
\( \frac{AC}{FG} = \frac{CD}{GH} = \frac{AD}{FH} \)
Thus, \( \frac {CD}{GH} = \frac {AC}{FG} \).
(ii) To show \( \triangle DCB \sim \triangle HGE \):
Consider \( \triangle DCB \) and \( \triangle HGE \).
1. \( \angle B = \angle E \) (Corresponding angles of similar triangles \( \triangle ABC \sim \triangle FEG \)).
2. \( \angle ACB = \angle FGE \) (Corresponding angles of similar triangles \( \triangle ABC \sim \triangle FEG \)).
Since CD bisects \( \angle ACB \) and GH bisects \( \angle FGE \):
\( \frac{1}{2}\angle ACB = \frac{1}{2}\angle FGE \)
\( \implies \angle DCB = \angle HGE \)
By the Angle-Angle (AA) similarity criterion,
\( \implies \triangle DCB \sim \triangle HGE \) (by AA similarity).
(iii) To show \( \triangle DCA \sim \triangle HGF \):
This is the same as part (i), just with a different order of vertices for \( \triangle ACD \). Let's re-verify.
Consider \( \triangle DCA \) and \( \triangle HGF \).
1. \( \angle DAC = \angle HFG \) (This is \( \angle A = \angle F \), corresponding angles of similar triangles \( \triangle ABC \sim \triangle FEG \)).
2. \( \angle ACD = \angle FGH \) (From part (i), half of \( \angle ACB \) and \( \angle FGE \) which are equal).
By the Angle-Angle (AA) similarity criterion,
\( \implies \triangle DCA \sim \triangle HGF \) (by AA similarity).
In simple words: We used the given fact that the two large triangles were similar. This meant their angles were equal, and their sides were in proportion. Because CD and GH split the angles in half, we could show that the smaller triangles formed (like \( \triangle ACD \) and \( \triangle FGH \)) also had equal angles, making them similar. This similarity then proved the side ratio and the similarity of other smaller triangle pairs.

Exam Tip: When dealing with angle bisectors in similar triangles, remember that the bisectors divide the corresponding angles into equal halves. This creates new pairs of equal angles, which are essential for proving the similarity of smaller triangles within the larger ones.

Question 11. In figure, E is a point on side CB produced on an isosceles triangle ABC with AB = AC. If AD \( \perp \) BC and EF \( \perp \) AC, Prove that \( \triangle ABD \sim \triangle ECF \). (CBSE 2012)
Answer: We are given an isosceles triangle \( \triangle ABC \) with \( AB = AC \). Point E is on CB extended. AD is perpendicular to BC (so AD is an altitude), and EF is perpendicular to AC (so EF is an altitude). We need to prove that \( \triangle ABD \sim \triangle ECF \).
Given:
1. \( \triangle ABC \) is an isosceles triangle with \( AB = AC \).
2. E is a point on BC produced.
3. AD \( \perp \) BC
4. EF \( \perp \) AC
To Prove: \( \triangle ABD \sim \triangle ECF \)
Proof:
Since \( AB = AC \) in \( \triangle ABC \), the angles opposite these sides are equal:
\( \angle B = \angle C \) (Angles opposite to equal sides in an isosceles triangle are equal).
So, \( \angle ABD = \angle ECF \).
Now, consider \( \triangle ABD \) and \( \triangle ECF \):
1. \( \angle ADB = 90^\circ \) (Since AD \( \perp \) BC).
2. \( \angle EFC = 90^\circ \) (Since EF \( \perp \) AC).
Therefore, \( \angle ADB = \angle EFC \) (each \( 90^\circ \)).
3. \( \angle ABD = \angle ECF \) (As proven above, \( \angle B = \angle C \)).
By the Angle-Angle (AA) similarity criterion, if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
\( \implies \triangle ABD \sim \triangle ECF \) (by AA similarity).
In simple words: We started with an isosceles triangle, so its base angles are equal. Then, we used the fact that AD and EF were altitudes, meaning they formed 90-degree angles. By finding two pairs of equal angles (the base angles and the 90-degree angles), we proved the two triangles were similar using the AA rule.

Exam Tip: In isosceles triangles, always remember that angles opposite equal sides are equal. Also, perpendicular lines or altitudes instantly provide \( 90^\circ \) angles, which are valuable for AA similarity proofs.

Question 12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of \( \triangle PQR \) (see figure). Show that \( \triangle ABC \sim \triangle PQR \).
Answer: We are given two triangles, \( \triangle ABC \) and \( \triangle PQR \). AD is a median of \( \triangle ABC \), and PM is a median of \( \triangle PQR \).
We are given that the sides AB, BC, and median AD of \( \triangle ABC \) are proportional to the sides PQ, QR, and median PM of \( \triangle PQR \).
So, \( \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM} \)
Since AD is the median to BC, D is the midpoint of BC. So, \( BD = DC = \frac{1}{2} BC \).
Since PM is the median to QR, M is the midpoint of QR. So, \( QM = MR = \frac{1}{2} QR \).
Now, let's rewrite the given proportionality using the median property:
\( \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM} \)
\( \frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{AD}{PM} \)
\( \implies \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM} \)
Now, consider \( \triangle ABD \) and \( \triangle PQM \).
The ratios of their corresponding sides are equal:
\( \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM} \)
By the Side-Side-Side (SSS) similarity criterion, if the corresponding sides of two triangles are proportional, then the triangles are similar.
\( \implies \triangle ABD \sim \triangle PQM \) (by SSS similarity).
Since these triangles are similar, their corresponding angles must be equal.
\( \implies \angle ABD = \angle PQM \)
This means \( \angle ABC = \angle PQR \).
Now, let's consider the main triangles, \( \triangle ABC \) and \( \triangle PQR \).
We are given that \( \frac{AB}{PQ} = \frac{BC}{QR} \) ........(1)
And we have proven that \( \angle ABC = \angle PQR \) ........(2)
By the Side-Angle-Side (SAS) similarity criterion, if two sides of one triangle are proportional to two sides of another triangle and the included angles are equal, then the two triangles are similar.
From (1) and (2), we can conclude that:
\( \implies \triangle ABC \sim \triangle PQR \) (by SAS similarity).
In simple words: We used the given side and median proportions to first prove that two smaller triangles (formed by the medians) were similar. This similarity showed that one pair of angles in the main triangles was equal. Then, combining this equal angle with the given proportional sides, we used the SAS (Side-Angle-Side) rule to show that the main triangles were also similar.

Exam Tip: When medians are involved in similarity proofs, remember that a median divides the opposite side into two equal parts. Use this fact to rewrite side ratios in terms of the median segments, which can help in establishing similarity for smaller triangles first.

Question 13. D is a point on the side BC of a triangle ABC such that \( \angle ADC = \angle BAC \). Show that \( CA^2 = CB \times CD \).
Answer: We have \( \triangle ABC \), and D is a point on side BC. We are given that \( \angle ADC = \angle BAC \). We need to show that \( CA^2 = CB \times CD \).
Consider \( \triangle ADC \) and \( \triangle BAC \).
1. \( \angle ADC = \angle BAC \) (This is given in the problem).
2. \( \angle ACD = \angle BCA \) (This is the common angle \( C \) shared by both triangles).
By the Angle-Angle (AA) similarity criterion, if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
\( \implies \triangle ADC \sim \triangle BAC \) (by AA similarity criterion).
Since these two triangles are similar, the ratios of their corresponding sides are equal:
\( \frac{AD}{BA} = \frac{DC}{AC} = \frac{AC}{BC} \)
From this proportionality, we take the part that involves AC, BC, and DC:
\( \frac{DC}{AC} = \frac{AC}{BC} \)
Now, cross-multiply these terms:
\( AC \times AC = BC \times DC \)
\( CA^2 = CB \times CD \)
Hence, proven.
In simple words: We identified two triangles, \( \triangle ADC \) and \( \triangle BAC \), and showed they were similar because they shared one common angle and had another pair of angles given as equal. Since similar triangles have proportional sides, we set up a ratio involving the sides we needed and cross-multiplied to get the desired equation.

Exam Tip: When proving relationships like \( CA^2 = CB \times CD \), it almost always involves proving similarity between two triangles and then using the proportionality of their sides. Make sure to correctly identify the corresponding vertices for the side ratios.

Question 14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that \( \triangle ABC \sim \triangle PQR \).
Answer: We are given two triangles, \( \triangle ABC \) and \( \triangle PQR \). AD is the median to BC, and PM is the median to QR.
We are given the proportionality:
\( \frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM} \) ........(1)
To show \( \triangle ABC \sim \triangle PQR \), we need to either prove another pair of sides are proportional (SSS) or prove that the included angle between the proportional sides is equal (SAS). We will use construction to achieve the latter.
**Construction:**
Produce AD to point E such that DE = AD. Join BE.
Produce PM to point N such that MN = PM. Join QN.

In \( \triangle ABD \) and \( \triangle ECD \):
\( AD = DE \) (by construction)
\( BD = CD \) (Since AD is a median)
\( \angle ADB = \angle CDE \) (Vertically opposite angles)
\( \implies \triangle ABD \cong \triangle ECD \) (by SAS congruence)
\( \implies AB = EC \) and \( \angle BAD = \angle CED \) (by CPCT)

Similarly, in \( \triangle PQM \) and \( \triangle NRM \):
\( PM = MN \) (by construction)
\( QM = RM \) (Since PM is a median)
\( \angle PMQ = \angle NMR \) (Vertically opposite angles)
\( \implies \triangle PQM \cong \triangle NRM \) (by SAS congruence)
\( \implies PQ = NR \) and \( \angle QPM = \angle MNR \) (by CPCT)

Now consider quadrilateral ABEC. Since diagonals AD and BE bisect each other (AD=DE and BD=CD), ABEC is a parallelogram.
\( \implies BE = AC \) (Opposite sides of a parallelogram are equal).
Similarly, for quadrilateral PQNR, since diagonals PM and QN bisect each other (PM=MN and QM=RM), PQNR is a parallelogram.
\( \implies QN = PR \) (Opposite sides of a parallelogram are equal).

From the given proportionality (1):
\( \frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM} \)
Multiply the AD/PM part by 2:
\( \frac{AB}{PQ} = \frac{AC}{PR} = \frac{2AD}{2PM} \)
Substitute \( 2AD = AE \) and \( 2PM = PN \):
\( \frac{AB}{PQ} = \frac{AC}{PR} = \frac{AE}{PN} \)
Now, substitute \( AC = BE \) and \( PR = QN \):
\( \frac{AB}{PQ} = \frac{BE}{QN} = \frac{AE}{PN} \)
This shows that the corresponding sides of \( \triangle ABE \) and \( \triangle PQN \) are proportional.
\( \implies \triangle ABE \sim \triangle PQN \) (by SSS similarity)
Since these triangles are similar, their corresponding angles are equal:
\( \implies \angle BAE = \angle QPN \) ........(2)
Similarly, by drawing CF (where F is on AB produced such that CF=AC) and RM (where M is on PQ produced such that RM=PR), it can be shown that \( \triangle ACE \sim \triangle PRN \).
\( \implies \angle CAE = \angle RPN \) ........(3)
Adding equations (2) and (3):
\( \angle BAE + \angle CAE = \angle QPN + \angle RPN \)
\( \angle BAC = \angle QPR \)
Now, consider \( \triangle ABC \) and \( \triangle PQR \).
We are given: \( \frac{AB}{PQ} = \frac{AC}{PR} \) ........(4)
We have proven: \( \angle BAC = \angle QPR \) ........(5)
By the Side-Angle-Side (SAS) similarity criterion, if two sides of one triangle are proportional to two sides of another triangle and the included angles are equal, then the two triangles are similar.
From (4) and (5), we conclude:
\( \implies \triangle ABC \sim \triangle PQR \) (by SAS similarity).
In simple words: We used a special construction to extend the medians, which helped us create parallelograms. This allowed us to prove two new, larger triangles were similar using the SSS rule. This similarity then showed that the angle between the given proportional sides in our original triangles was equal. Finally, with the proportional sides and the equal included angle, we proved the main triangles were similar using the SAS rule.

Exam Tip: When medians are involved and you need to prove SAS similarity, consider extending the medians to create a parallelogram. This construction allows you to relate the given proportional sides and medians to an included angle for the SAS criterion.

Question 15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Answer: Let \( \triangle ABC \) represent the pole and its shadow, and \( \triangle DEF \) represent the tower and its shadow.
Given:
Height of the pole (AB) = 6 m
Length of the pole's shadow (BC) = 4 m
Length of the tower's shadow (EF) = 28 m
Let the height of the tower (DE) be \( h \) m.
At the same time of day, the angle of elevation of the sun is the same for both the pole and the tower. This means the angle the sun's rays make with the ground is identical.
Consider \( \triangle ABC \) (for the pole) and \( \triangle DEF \) (for the tower).
1. \( \angle ABC = \angle DEF = 90^\circ \) (Both the pole and the tower are vertical to the ground).
2. \( \angle BAC = \angle EDF \) (Angle of elevation of the sun at the same time).
By the Angle-Angle (AA) similarity criterion, if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
\( \implies \triangle ABC \sim \triangle DEF \) (by AA similarity).
Since these triangles are similar, the ratios of their corresponding sides are equal:
\( \frac{AB}{DE} = \frac{BC}{EF} \)
Substitute the known values:
\( \frac{6}{h} = \frac{4}{28} \)
Now, we solve for \( h \):
\( 6 \times 28 = 4 \times h \)
\( h = \frac{6 \times 28}{4} \)
\( h = 6 \times 7 \)
\( h = 42 \)
Therefore, the height of the tower is 42 m.
In simple words: We used the idea that the sun's angle in the sky is the same for both the pole and the tower. This means the triangles formed by each object and its shadow are similar. Since they are similar, their heights and shadow lengths are proportional. We set up a ratio and solved for the unknown tower height.

Exam Tip: Problems involving shadows and heights of objects often use the concept of similar triangles due to the constant angle of elevation of the sun. Always draw a clear diagram and identify corresponding sides and angles correctly for setting up the proportion.

Question 16. If AD and PM are medians of triangles \( \triangle ABC \) and \( \triangle PQR \), respectively where \( \triangle ABC \sim \triangle PQR \). Prove that: \( \frac{AB}{PQ}=\frac{AD}{PM} \).
Answer: We are given that AD and PM are medians of \( \triangle ABC \) and \( \triangle PQR \) respectively. We are also given that \( \triangle ABC \sim \triangle PQR \). We need to prove that \( \frac{AB}{PQ}=\frac{AD}{PM} \).
Since \( \triangle ABC \sim \triangle PQR \), the ratios of their corresponding sides are equal, and their corresponding angles are equal:
\( \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} \) ........(1)
And \( \angle A = \angle P \), \( \angle B = \angle Q \), \( \angle C = \angle R \).
Now, consider \( \triangle ABD \) and \( \triangle PQM \).
1. \( \angle B = \angle Q \) (Corresponding angles of similar triangles \( \triangle ABC \sim \triangle PQR \)).
2. We know that D is the midpoint of BC (since AD is a median), so \( BD = \frac{1}{2} BC \).
Similarly, M is the midpoint of QR (since PM is a median), so \( QM = \frac{1}{2} QR \).
From equation (1), we have \( \frac{AB}{PQ} = \frac{BC}{QR} \).
We can write this as \( \frac{AB}{PQ} = \frac{2BD}{2QM} \).
\( \implies \frac{AB}{PQ} = \frac{BD}{QM} \) ........(2)
By the Side-Angle-Side (SAS) similarity criterion, if two sides of one triangle are proportional to two sides of another triangle and the included angles are equal, then the two triangles are similar.
From (2) and \( \angle B = \angle Q \), we can conclude that:
\( \implies \triangle ABD \sim \triangle PQM \) (by SAS similarity).
Since these triangles are similar, the ratios of their corresponding sides are equal:
\( \implies \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM} \)
From this, we get the required result:
\( \frac{AB}{PQ}=\frac{AD}{PM} \)
Hence, proven.
In simple words: We used the given similarity of the large triangles to establish that one pair of angles and two pairs of sides were proportional. Because AD and PM were medians, they split the base sides in half. This helped us set up a side ratio for the smaller triangles involving the medians. Using this ratio and the equal angles, we proved the smaller triangles were similar, which then directly showed the proportion between the side AB, PM, and the medians AD, PM.

Exam Tip: When given that two triangles are similar and also involve medians, remember to use both facts: proportionality of all corresponding sides/angles from similarity and the property of medians bisecting sides. This allows for proving similarity of smaller triangles and establishing the desired ratio involving medians.