Samacheer Kalvi Class 10 Maths Solutions Chapter 4 Geometry Exercise 4.2

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.2

Question 1. In ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC
(i) If \( \frac { AD }{ DB } = \frac { 3 }{ 4 } \) and AC = 15 cm find AE.
(ii) If AD = 8x – 7, DB = 5x – 3, AE = 4x – 3 and EC = 3x – 1, find the value of x.
Answer:
(i) Given: In ∆ABC, DE || BC, \( \frac { AD }{ DB } = \frac { 3 }{ 4 } \), AC = 15 cm.
We need to find AE.
According to the Basic Proportionality Theorem (Thales Theorem), if a line is drawn parallel to one side of a triangle intersecting the other two sides, then the other two sides are divided in the same ratio. This theorem is fundamental for solving problems involving parallel lines in triangles.
Since DE || BC, we have:
\( \frac { AD }{ DB } = \frac { AE }{ EC } \)
Substitute the given values:
\( \frac { 3 }{ 4 } = \frac { AE }{ EC } \)
Let AE = x. Then EC = AC - AE = 15 - x.
So, \( \frac { 3 }{ 4 } = \frac { x }{ 15-x } \)
Now, cross-multiply to solve for x:
\( 3(15-x) = 4x \)
\( 45 - 3x = 4x \)
Add 3x to both sides:
\( 45 = 4x + 3x \)
\( 45 = 7x \)
Divide by 7:
\( x = \frac { 45 }{ 7 } \)
\( x \approx 6.43 \)
Therefore, AE = 6.43 cm (approximately).

(ii) Given: In ∆ABC, DE || BC.
AD = 8x - 7, DB = 5x - 3, AE = 4x - 3, EC = 3x - 1.
We need to find the value of x.
By the Basic Proportionality Theorem (BPT), since DE || BC, the ratio of the segments on the sides AB and AC must be equal.
\( \frac { AD }{ DB } = \frac { AE }{ EC } \)
Substitute the given expressions for the segments:
\( \frac { 8x-7 }{ 5x-3 } = \frac { 4x-3 }{ 3x-1 } \)
Now, cross-multiply to solve for x:
\( (8x-7)(3x-1) = (4x-3)(5x-3) \)
Expand both sides:
\( 24x^2 - 8x - 21x + 7 = 20x^2 - 12x - 15x + 9 \)
Combine like terms on each side:
\( 24x^2 - 29x + 7 = 20x^2 - 27x + 9 \)
Move all terms to one side to form a quadratic equation:
\( 24x^2 - 20x^2 - 29x + 27x + 7 - 9 = 0 \)
\( 4x^2 - 2x - 2 = 0 \)
Divide the entire equation by 2 to simplify:
\( 2x^2 - x - 1 = 0 \)
Now, factor the quadratic equation. We look for two numbers that multiply to \( 2 \times -1 = -2 \) and add to -1. These numbers are -2 and 1.
\( 2x^2 - 2x + x - 1 = 0 \)
Factor by grouping:
\( 2x(x - 1) + 1(x - 1) = 0 \)
\( (x - 1)(2x + 1) = 0 \)
Set each factor to zero to find the possible values of x:
\( x - 1 = 0 \quad \text{or} \quad 2x + 1 = 0 \)
\( x = 1 \quad \text{or} \quad 2x = -1 \)
\( x = 1 \quad \text{or} \quad x = -\frac { 1 }{ 2 } \)
Since lengths cannot be negative, we must omit the negative value. A physical length must always be positive.
If \( x = -\frac { 1 }{ 2 } \), then \( DB = 5(-\frac { 1 }{ 2 }) - 3 = -2.5 - 3 = -5.5 \), which is not possible.
Therefore, the value of x is 1.
In simple words: When a line inside a triangle is parallel to one side, it cuts the other two sides into parts that have the same proportions. We use this rule to set up an equation with the given lengths and solve for 'x'. We ignore any 'x' value that would make a side length negative.

🎯 Exam Tip: Always check if any derived value of 'x' would result in a negative length. Lengths in geometry must always be positive. If an 'x' value leads to a negative length, it should be discarded.

Question 2. ABCD is a trapezium in which AB || DC and P,Q are points on AD and BC respectively, such that PQ || DC if PD = 18 cm, BQ = 35 cm, QC = 15 cm and AP = x, find the value of x.
Answer:
Given: ABCD is a trapezium with AB || DC. P is on AD, Q is on BC, and PQ || DC.
Also given: PD = 18 cm, BQ = 35 cm, QC = 15 cm, AP = x.
We need to find the value of x.
First, let's draw a diagonal AC which intersects PQ at point S. This helps us apply the Basic Proportionality Theorem in two different triangles.

Since AB || DC and PQ || DC, it means PQ || AB as well.
Consider ∆ABC:
In ∆ABC, QS || AB (since PQ || AB).
By the Basic Proportionality Theorem, we have:
\( \frac { AS }{ SC } = \frac { BQ }{ QC } \)
Substitute the given values for BQ and QC:
\( \frac { AS }{ SC } = \frac { 35 }{ 15 } \)
Simplify the ratio:
\( \frac { AS }{ SC } = \frac { 7 }{ 3 } \) .....(1)
Now, consider ∆ACD:
In ∆ACD, PS || DC (since PQ || DC).
By the Basic Proportionality Theorem, we have:
\( \frac { AS }{ SC } = \frac { AP }{ PD } \)
Substitute the given values for AP and PD:
\( \frac { AS }{ SC } = \frac { x }{ 18 } \) .....(2)
From equations (1) and (2), since both ratios are equal to \( \frac { AS }{ SC } \), we can equate them:
\( \frac { 7 }{ 3 } = \frac { x }{ 18 } \)
Now, solve for x:
Multiply both sides by 18:
\( x = \frac { 7 }{ 3 } \times 18 \)
\( x = 7 \times 6 \)
\( x = 42 \)
So, AP = 42 cm.
The question asks for the value of x, which is 42.
We can also find the length of AD: AD = AP + PD = x + 18 = 42 + 18 = 60 cm.
In simple words: We draw a diagonal in the trapezium to create two triangles. Using the rule that parallel lines divide sides proportionally in each triangle, we get two ratios that are equal. By setting these ratios equal to each other, we can find the unknown length 'x'.

🎯 Exam Tip: When dealing with trapeziums and parallel lines, drawing a diagonal can help you apply the Basic Proportionality Theorem in separate triangles, which simplifies the problem greatly.

Question 3. In ∆ABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE || BC.
(i) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm.
(ii) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm.
Answer:
For DE to be parallel to BC, by the converse of the Basic Proportionality Theorem, the line segment DE must divide the sides AB and AC in the same ratio. That means, \( \frac { AD }{ DB } \) must be equal to \( \frac { AE }{ EC } \).
(i) Given: AB = 12 cm, AD = 8 cm, AE = 12 cm, AC = 18 cm.
First, find the length of DB:
DB = AB - AD = 12 cm - 8 cm = 4 cm.
Next, find the length of EC:
EC = AC - AE = 18 cm - 12 cm = 6 cm.
Now, calculate the ratios:
\( \frac { AD }{ DB } = \frac { 8 }{ 4 } = 2 \)
\( \frac { AE }{ EC } = \frac { 12 }{ 6 } = 2 \)
Since \( \frac { AD }{ DB } = \frac { AE }{ EC } \) (both are equal to 2), by the converse of the Basic Proportionality Theorem, DE || BC.

(ii) Given: AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm, AE = 1.8 cm.
First, find the length of DB:
DB = AB - AD = 5.6 cm - 1.4 cm = 4.2 cm.
Next, find the length of EC:
EC = AC - AE = 7.2 cm - 1.8 cm = 5.4 cm.
Now, calculate the ratios:
\( \frac { AD }{ DB } = \frac { 1.4 }{ 4.2 } = \frac { 14 }{ 42 } = \frac { 1 }{ 3 } \)
\( \frac { AE }{ EC } = \frac { 1.8 }{ 5.4 } = \frac { 18 }{ 54 } = \frac { 1 }{ 3 } \)
Since \( \frac { AD }{ DB } = \frac { AE }{ EC } \) (both are equal to \( \frac { 1 }{ 3 } \)), by the converse of the Basic Proportionality Theorem, DE || BC.

In simple words: To show that line DE is parallel to side BC of a triangle, we need to check if DE divides the other two sides (AB and AC) into parts that have the same ratio. If the ratio of the top part to the bottom part on one side is the same as on the other side, then the lines are parallel.

🎯 Exam Tip: Remember the "converse" of a theorem. For the Basic Proportionality Theorem, if the line divides two sides in the same ratio, then the line is parallel to the third side. This is crucial for proofs.

Question 4. In fig. if PQ || BC and PR || CD prove that
(i) \( \frac { AR }{ AD } = \frac { AQ }{ AB } \)
(ii) \( \frac { QB }{ AQ } = \frac { DR }{ AR } \)
Answer:
Given: In the figure, PQ || BC and PR || CD.
We need to prove the given ratio relations.

(i) To prove: \( \frac { AR }{ AD } = \frac { AQ }{ AB } \)
Consider ∆ABC:
Given that PQ || BC.
By the Basic Proportionality Theorem (BPT), if a line parallel to one side of a triangle intersects the other two sides, then it divides the two sides proportionally. This theorem is also known as Thales's Theorem.
So, in ∆ABC, since PQ || BC, we have:
\( \frac { AQ }{ AB } = \frac { AP }{ AC } \) .....(1)
Now, consider ∆ACD:
Given that PR || CD.
By the Basic Proportionality Theorem, in ∆ACD, since PR || CD, we have:
\( \frac { AP }{ AC } = \frac { AR }{ AD } \) .....(2)
From (1) and (2), we can equate the expressions that are both equal to \( \frac { AP }{ AC } \):
\( \frac { AQ }{ AB } = \frac { AR }{ AD } \)
Thus, the first part is proved.
(ii) To prove: \( \frac { QB }{ AQ } = \frac { DR }{ AR } \)
From part (i), we already proved that \( \frac { AQ }{ AB } = \frac { AR }{ AD } \).
We can also write this as: \( \frac { AB }{ AQ } = \frac { AD }{ AR } \)
Now, subtract 1 from both sides:
\( \frac { AB }{ AQ } - 1 = \frac { AD }{ AR } - 1 \)
Find a common denominator for each side:
\( \frac { AB - AQ }{ AQ } = \frac { AD - AR }{ AR } \)
We know that AB - AQ = QB and AD - AR = DR.
Substitute these into the equation:
\( \frac { QB }{ AQ } = \frac { DR }{ AR } \)
Thus, the second part is also proved.
In simple words: We use the rule that a line parallel to one side of a triangle divides the other two sides into parts that are in the same proportion. By applying this rule to two different triangles that share a common side or point, we can link the ratios and prove the required relationships.

🎯 Exam Tip: When given multiple parallel lines in a complex figure, break down the problem into simpler triangles. Apply the BPT (Basic Proportionality Theorem) to each triangle separately and then combine the results using shared ratios.

Question 5. Rhombus PQRB is inscribed in ∆ABC such that ∠B is one of its angle. P, Q and R lie on AB, AC and BC respectively. If AB = 12 cm and BC = 6 cm, find the sides PQ, RB of the rhombus.
Answer:
Given: ∆ABC with a rhombus PQRB inscribed. ∠B is an angle of the rhombus. P lies on AB, Q on AC, and R on BC.
AB = 12 cm, BC = 6 cm.
We need to find the side length of the rhombus (PQ, RB).
Let the side length of the rhombus be 'x' cm. So, PB = BR = PQ = QR = x.
Since PQRB is a rhombus, PQ || BR (which is part of BC). So, PQ || BC.

Since PQ || BC, by the Basic Proportionality Theorem in ∆ABC, we have:
\( \frac { AP }{ PB } = \frac { AQ }{ QC } \)
Also, from similar triangles ∆APQ and ∆ABC (since PQ || BC), the ratio of their corresponding sides is equal.
\( \frac { AP }{ AB } = \frac { PQ }{ BC } \)
We know AB = 12 cm, BC = 6 cm, and PQ = x. Also, AP = AB - PB = 12 - x.
Substitute these values into the ratio:
\( \frac { 12-x }{ 12 } = \frac { x }{ 6 } \)
Now, solve for x:
Multiply both sides by 12:
\( 12 \left( \frac { 12-x }{ 12 } \right) = 12 \left( \frac { x }{ 6 } \right) \)
\( 12 - x = 2x \)
Add x to both sides:
\( 12 = 2x + x \)
\( 12 = 3x \)
Divide by 3:
\( x = \frac { 12 }{ 3 } \)
\( x = 4 \)
So, the side of the rhombus is 4 cm.
Therefore, PQ = RB = 4 cm.
In simple words: We know the rhombus has equal sides. Since one side of the rhombus is parallel to a side of the larger triangle, we can use the property of similar triangles. By setting up a ratio of the sides, we can find the unknown side length 'x' of the rhombus.

🎯 Exam Tip: When a shape (like a rhombus or square) is inscribed in a triangle with one side parallel to the base, remember to use similar triangles to set up proportional relationships between their sides. This is a common technique for such problems.

Question 6. In trapezium ABCD, AB || DC, E and F are points on non-parallel sides AD and BC respectively, such that EF || AB. Show that \( \frac { AE }{ ED } = \frac { BF }{ FC } \)
Answer:
Given: ABCD is a trapezium with AB || DC. E is a point on AD and F is a point on BC, such that EF || AB.
We need to show that \( \frac { AE }{ ED } = \frac { BF }{ FC } \).

Construction: Join the diagonal AC. Let AC intersect EF at point P.
Proof:
Since AB || DC and EF || AB, it implies that EF || DC as well (lines parallel to the same line are parallel to each other). This is a helpful property when dealing with multiple parallel lines.
Consider ∆ABC:
In ∆ABC, PF || AB (since EF || AB, and P is on EF).
By the Basic Proportionality Theorem (BPT), we have:
\( \frac { AP }{ PC } = \frac { BF }{ FC } \) .....(1)
Now, consider ∆ADC:
In ∆ADC, PE || DC (since EF || DC, and P is on EF).
By the Basic Proportionality Theorem (BPT), we have:
\( \frac { AP }{ PC } = \frac { AE }{ ED } \) .....(2)
From equations (1) and (2), since both ratios are equal to \( \frac { AP }{ PC } \), we can equate them:
\( \frac { AE }{ ED } = \frac { BF }{ FC } \)
Thus, it is proved.
In simple words: When you have a shape with two parallel sides (a trapezium) and another line inside it that is also parallel to those sides, you can draw a diagonal to split the shape into two triangles. Then, using the rule that parallel lines divide the sides of a triangle proportionally, you can show that the segments on the non-parallel sides have equal ratios.

🎯 Exam Tip: When proving proportionality in trapeziums, always draw a diagonal to create two triangles. This allows you to apply the Basic Proportionality Theorem (BPT) twice and then link the results to reach your desired proof.

Question 7. In figure DE || BC and CD || EF Prove that AD\(^2\) = AB × AF.
Answer:
Given: In the figure, DE || BC and CD || EF.
We need to prove that AD\(^2\) = AB × AF.

Proof:
Consider ∆ABC:
Given that DE || BC. D is on AB and E is on AC. This is a common setup for Thales's Theorem, where a line segment connects two sides of a triangle and is parallel to the third side.
By the Basic Proportionality Theorem (BPT), we have:
\( \frac { AB }{ AD } = \frac { AC }{ AE } \) .....(1)
Now, consider ∆ADC:
Given that EF || DC. F is on AD and E is on AC.
By the Basic Proportionality Theorem (BPT), we have:
\( \frac { AD }{ AF } = \frac { AC }{ AE } \) .....(2)
From equations (1) and (2), both ratios are equal to \( \frac { AC }{ AE } \). Therefore, we can equate them:
\( \frac { AB }{ AD } = \frac { AD }{ AF } \)
Now, cross-multiply:
\( AD \times AD = AB \times AF \)
\( AD^2 = AB \times AF \)
Thus, the relationship is proved.
In simple words: We have two sets of parallel lines inside a larger triangle. By using the Basic Proportionality Theorem twice – once for each pair of parallel lines within its respective triangle – we get two ratios that share a common part. By equating these common ratios, we can find the required relationship between the lengths.

🎯 Exam Tip: When faced with multiple parallel lines within a single figure, clearly identify the triangle for each application of the Basic Proportionality Theorem. Ensure you correctly match the corresponding sides in each ratio.

Question 8. In ∆ABC, AD is the bisector of ∠A meeting side BC at D, if AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.
Answer:
Given: In ∆ABC, AD is the angle bisector of ∠A. D is on BC.
AB = 10 cm, AC = 14 cm, BC = 6 cm.
We need to find the lengths of BD and DC.
Let BD = x cm. Then, since BC = 6 cm, DC = (6 - x) cm.

By the Angle Bisector Theorem, the bisector of an angle in a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle. This theorem is a powerful tool for finding unknown side lengths.
So, in ∆ABC, since AD is the angle bisector of ∠A, we have:
\( \frac { BD }{ DC } = \frac { AB }{ AC } \)
Substitute the known values:
\( \frac { x }{ 6-x } = \frac { 10 }{ 14 } \)
Simplify the ratio \( \frac { 10 }{ 14 } \) to \( \frac { 5 }{ 7 } \):
\( \frac { x }{ 6-x } = \frac { 5 }{ 7 } \)
Now, cross-multiply to solve for x:
\( 7x = 5(6-x) \)
\( 7x = 30 - 5x \)
Add 5x to both sides:
\( 7x + 5x = 30 \)
\( 12x = 30 \)
Divide by 12:
\( x = \frac { 30 }{ 12 } \)
Simplify the fraction:
\( x = \frac { 5 }{ 2 } \)
\( x = 2.5 \)
So, BD = 2.5 cm.
Now, find DC:
DC = 6 - x = 6 - 2.5 = 3.5 cm.
Thus, BD = 2.5 cm and DC = 3.5 cm.
In simple words: When a line splits an angle of a triangle and meets the opposite side, it divides that side into two pieces. The lengths of these pieces are in the same ratio as the lengths of the other two sides of the triangle. We use this rule to set up an equation and find the lengths of the two pieces.

🎯 Exam Tip: Clearly identify the angle bisector and the sides it divides. Ensure you correctly set up the proportion with the side adjacent to the divided segment on the numerator and the opposite side on the denominator, i.e., \( \frac{BD}{DC} = \frac{AB}{AC} \).

Question 9. Check whether AD is bisector of ∠A of ∆ABC in each of the following,
(i) AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm.
(ii) AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm.
Answer:
For AD to be the bisector of ∠A, by the converse of the Angle Bisector Theorem, the ratio of the sides \( \frac { AB }{ AC } \) must be equal to the ratio of the segments \( \frac { BD }{ DC } \).
(i) Given: AB = 5 cm, AC = 10 cm, BD = 1.5 cm, CD = 3.5 cm.
Calculate the ratio \( \frac { BD }{ DC } \):
\( \frac { BD }{ DC } = \frac { 1.5 }{ 3.5 } = \frac { 15 }{ 35 } = \frac { 3 }{ 7 } \)
Calculate the ratio \( \frac { AB }{ AC } \):
\( \frac { AB }{ AC } = \frac { 5 }{ 10 } = \frac { 1 }{ 2 } \)
Since \( \frac { BD }{ DC } = \frac { 3 }{ 7 } \) and \( \frac { AB }{ AC } = \frac { 1 }{ 2 } \), these ratios are not equal. \( \frac { 3 }{ 7 } \neq \frac { 1 }{ 2 } \).
Therefore, AD is not a bisector of ∠A.

(ii) Given: AB = 4 cm, AC = 6 cm, BD = 1.6 cm, CD = 2.4 cm.
Calculate the ratio \( \frac { BD }{ DC } \):
\( \frac { BD }{ DC } = \frac { 1.6 }{ 2.4 } = \frac { 16 }{ 24 } = \frac { 2 }{ 3 } \)
Calculate the ratio \( \frac { AB }{ AC } \):
\( \frac { AB }{ AC } = \frac { 4 }{ 6 } = \frac { 2 }{ 3 } \)
Since \( \frac { BD }{ DC } = \frac { 2 }{ 3 } \) and \( \frac { AB }{ AC } = \frac { 2 }{ 3 } \), these ratios are equal. \( \frac { BD }{ DC } = \frac { AB }{ AC } \).
Therefore, by the converse of the Angle Bisector Theorem, AD is the internal bisector of ∠A.

In simple words: To check if a line segment inside a triangle is an angle bisector, we see if it divides the opposite side in the same proportion as the other two sides of the triangle. If the ratios match, it's an angle bisector; otherwise, it's not.

🎯 Exam Tip: The converse of the Angle Bisector Theorem is used to *verify* if a given line is an angle bisector. Always calculate both ratios \( \frac{BD}{DC} \) and \( \frac{AB}{AC} \) and compare them directly for equality.

Question 10. In figure ∠QPR = 90°, PS is its bisector. If ST \( \perp \) PR, prove that ST × (PQ + PR) = PQ × PR.
Answer:
Given: In ∆PQR, ∠QPR = 90°. PS is the bisector of ∠P. ST \( \perp \) PR (T is on PR).
We need to prove that ST × (PQ + PR) = PQ × PR.

Proof:
In ∆PQR, PS is the angle bisector of ∠P.
By the Angle Bisector Theorem, we have:
\( \frac { PQ }{ PR } = \frac { QS }{ SR } \) (This is a common application of the theorem, relating the sides to the segments of the opposite side)
Now, let's rearrange this to get a different form for comparison. Add 1 to both sides:
\( \frac { PQ }{ PR } + 1 = \frac { QS }{ SR } + 1 \)
\( \frac { PQ + PR }{ PR } = \frac { QS + SR }{ SR } \)
Since QS + SR = QR (the whole side), we get:
\( \frac { PQ + PR }{ PR } = \frac { QR }{ SR } \) .....(1)
Next, consider ∆RST and ∆RQP.
We have ∠SRT = ∠QRP = ∠R (Common angle).
Also, ∠RTS = 90° (since ST \( \perp \) PR, so ST is perpendicular to the line PR).
And ∠RQP = 90° (given ∠QPR = 90°, so the angle at Q in ∆RQP is also 90° since it is a right-angled triangle). Wait, ∠QPR=90 means angle P is 90 degrees, not Q. So ∆PQR is a right-angled triangle at P. Let's re-evaluate the similar triangles section. If ∠QPR = 90°, then P is the right angle. So, in ∆RQP, ∠P = 90°. In ∆RST, ∠T = 90° (given ST \( \perp \) PR). We have ∠R (common to both ∆RST and ∆RQP). Therefore, by AA similarity criterion, ∆RST ~ ∆RQP.
From the similarity of triangles, the ratios of corresponding sides are equal:
\( \frac { ST }{ QP } = \frac { RS }{ RQ } \)
Rearranging this, we get:
\( \frac { ST }{ PQ } = \frac { SR }{ QR } \)
Now, take the reciprocal of both sides:
\( \frac { PQ }{ ST } = \frac { QR }{ SR } \) .....(2)
From (1) and (2), both expressions are equal to \( \frac { QR }{ SR } \), so we can equate them:
\( \frac { PQ + PR }{ PR } = \frac { PQ }{ ST } \)
Finally, cross-multiply to get the desired result:
\( ST \times (PQ + PR) = PQ \times PR \)
Thus, the proof is complete.
In simple words: We start by using the angle bisector rule in the main triangle. Then, we identify two smaller triangles that are similar because they share an angle and both have a right angle. By using the equal side ratios from these similar triangles and combining it with our first rule, we can prove the given equation.

🎯 Exam Tip: When proving complex geometric identities, look for opportunities to use both the Angle Bisector Theorem and triangle similarity. Sometimes, rearranging a ratio or adding/subtracting 1 can transform it into a more useful form for comparison with other ratios.

Question 11. ABCD is a quadrilateral in which AB = AD, the bisector of ∠BAC and ∠CAD intersect the sides BC and CD at the points E and F respectively. Prove that EF || BD.
Answer:
Given: ABCD is a quadrilateral. AB = AD. AE is the bisector of ∠BAC (E on BC). AF is the bisector of ∠CAD (F on CD).
We need to prove that EF || BD.

Proof:
Consider ∆ABC:
AE is the internal bisector of ∠BAC (given). The Angle Bisector Theorem states that an angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle.
So, by the Angle Bisector Theorem in ∆ABC:
\( \frac { BE }{ EC } = \frac { AB }{ AC } \) .....(1)
Now, consider ∆ADC:
AF is the internal bisector of ∠CAD (given).
So, by the Angle Bisector Theorem in ∆ADC:
\( \frac { DF }{ FC } = \frac { AD }{ AC } \) .....(2)
We are given that AB = AD.
Substitute AB for AD in equation (2):
\( \frac { DF }{ FC } = \frac { AB }{ AC } \) .....(3)
From equations (1) and (3), both ratios are equal to \( \frac { AB }{ AC } \). Therefore, we can equate them:
\( \frac { BE }{ EC } = \frac { DF }{ FC } \)
Now, consider ∆BCD:
E is a point on BC and F is a point on CD.
We have shown that \( \frac { BE }{ EC } = \frac { DF }{ FC } \).
By the converse of the Basic Proportionality Theorem (also known as Thales's Theorem), if a line divides two sides of a triangle proportionally, then the line is parallel to the third side. This theorem is crucial for establishing parallel lines.
Since EF divides the sides BC and CD of ∆BCD proportionally, it implies that EF || BD.
Thus, the proof is complete.
In simple words: We use the angle bisector rule in two different triangles, ∆ABC and ∆ADC. Since we know AB = AD, the ratios \( \frac { BE }{ EC } \) and \( \frac { DF }{ FC } \) both become equal to \( \frac { AB }{ AC } \). Because these ratios are equal, it means that in the triangle ∆BCD, the line EF divides two of its sides proportionally, which proves that EF is parallel to the third side, BD.

🎯 Exam Tip: When given an isosceles condition (like AB=AD), remember to leverage it to simplify ratios. The converse of the Basic Proportionality Theorem is a key tool for proving that two lines are parallel.

Question 12. Construct a ∆PQR which the base PQ = 4.5 cm, R = 35° and the median from R to RG is 6 cm.
Answer:
Rough Diagram:

Main Construction Diagram:

Steps of construction:
1. Draw a line segment PQ = 4.5 cm.
2. At P, draw a ray PE such that ∠QPE = 35°. This ray forms the desired angle at P, which helps in defining the triangle's shape.
3. At P, draw another ray PF perpendicular to PE (∠EPF = 90°). This line will guide us to find the center of the circumcircle.
4. Draw the perpendicular bisector to PQ. This bisector will intersect PF at point O and PQ at point G. Point O is the circumcenter of ∆PQR. The perpendicular bisector ensures that O is equidistant from P and Q.
5. With O as the center and OP as the radius, draw a circle. This circle will pass through P and Q, and the vertex R will lie on this circle.
6. From G (the midpoint of PQ), mark arcs of radius 6 cm (the length of the median RG) on the circle. Mark the intersection points as R and S. The median from R to PQ is RG, so R is 6cm from G. RG is not perpendicular to PQ.
7. Join PR and RQ. ∆PQR is the required triangle.
In simple words: We first draw the base of the triangle. Then, we use the given angle and properties of a circumcircle to find where the third vertex R should be. We draw a circle that passes through the base points and then find point R on this circle by using the given length of the median from R to the midpoint of the base.

🎯 Exam Tip: When constructing triangles with a given base, vertical angle, and median, remember that the vertical angle lies on the circumference of a segment of a circle whose chord is the base. The median length is used to locate the exact position of the third vertex on that circle segment.

Question 13. Construct a ΔPQR in which QR = 5 cm, ∠P = 40° and the median PG from P to QR is 4.4 cm. Find the length of the altitude from P to QR.
Answer:

Rough Diagram

Steps of construction:
1. First, draw a line segment QR with a length of 5 cm.
2. At point Q, draw a ray QE such that \( \angle RQE = 40^\circ \). This angle helps to define the position of point P.
3. At point Q, draw another ray QF such that \( \angle EQF = 90^\circ \). This line will be used to find the center of the circle.
4. Draw the perpendicular bisector of the line segment QR. This bisector will intersect the ray QF at a point O and the line segment QR at a point G. The perpendicular bisector ensures symmetry.
5. Using O as the center and OQ as the radius, draw a circle. Point P will lie on this circle.
6. From point G on QR, mark arcs on the circle with a radius of 4.4 cm (the length of the median PG). Label these points as P and S. This step fixes the position of P based on the median length.
7. Join the points P to R and P to Q to complete the triangle PQR.
8. From point P, draw a line segment PN perpendicular to QR. This line PN represents the altitude from P to QR and meets QR at point N.
9. Measure the length of the altitude PN. This will give the required length.
PN = 2.2 cm.
In simple words: First, draw the base QR. Then, use the given angle to start drawing the triangle. Find the center of a circle that passes through P, Q, and R. Mark point P using the median length and draw the triangle. Finally, draw a straight line from P down to QR at a right angle to find the height, and measure it.

🎯 Exam Tip: In construction problems, always start with a clear rough diagram to visualize the steps. Use a sharp pencil and precise measurements for accuracy, especially when drawing arcs and bisectors.

Question 14. Construct a ΔPQR such that QR = 6.5 cm, ∠P = 60° and the altitude from P to QR is of length 4.5 cm.
Answer:

Rough Diagram

Steps of construction:
1. Draw a line segment QR with a length of 6.5 cm. This forms the base of the triangle.
2. At point Q, draw a ray QE such that \( \angle RQE = 60^\circ \). This angle is related to the vertical angle \( \angle P \).
3. At point Q, draw another ray QF such that \( \angle EQF = 90^\circ \). This line will help in finding the center for the circumcircle.
4. Draw the perpendicular bisector of the line segment QR. This bisector will intersect QF at point O and QR at point G. Point O is the center of the circle.
5. Using O as the center and OQ as the radius, draw a circle. Point P will lie on this circle.
6. Draw a line XY parallel to QR at a distance of 4.5 cm (the given altitude length) from QR. To do this, mark a point M on the perpendicular bisector such that GM = 4.5 cm, then draw line XY through M parallel to QR. This line contains all possible points for P that have the required altitude.
7. The line XY will intersect the circle at points P and S. Both P and S can be the vertex of the triangle.
8. Join PR and PQ. The triangle PQR is the required triangle.
In simple words: Start by drawing the base QR. Use the angle \( \angle P \) to guide the shape of the triangle by constructing a related angle at Q and finding the center of a circle. Then, draw a line parallel to the base at the exact height (altitude) given. Where this height line crosses the circle, that's where the top point of your triangle goes. Finally, connect the points to make the triangle.

🎯 Exam Tip: When constructing triangles with a given angle and altitude, remember that the vertex opposite the base lies on a line parallel to the base at the altitude distance and also on the arc of a circle. Both conditions must be met for a successful construction.

Question 15. Construct a ΔABC such that AB = 5.5 cm, ∠C = 25° and the altitude from C to AB is 4 cm.
Answer:

Rough Diagram

Steps of construction:
1. Draw a line segment AB of length 5.5 cm. This will be the base of the triangle.
2. At point A, draw a ray AE such that \( \angle BAE = 25^\circ \). This angle helps to define the shape based on \( \angle C \).
3. At point A, draw another ray AF such that \( \angle EAF = 90^\circ \). This ray is crucial for finding the center of the circumcircle.
4. Draw the perpendicular bisector of the line segment AB. This bisector will intersect the ray AF at point O and the line segment AB at point G. Point O will be the center of the circle.
5. Using O as the center and OB as the radius, draw a circle. Point C will lie on this circle.
6. Draw a line XY parallel to AB at a distance of 4 cm (the given altitude length) from AB. Mark a point M on the perpendicular bisector such that GM = 4 cm. Then draw line XY through M parallel to AB. This line contains all possible positions for vertex C at the correct height.
7. The line XY will intersect the circle at points C and D. Both C and D can be the third vertex.
8. Join AC and BC to form the required triangle ABC.
In simple words: First, draw the base AB. Then, use the given angle at C to find the right location for point C by drawing a special angle at A and a perpendicular line. Find the center of a circle that passes through A, B, and C. Next, draw a line parallel to the base at the specified height. Where this height line crosses the circle, that is where point C will be. Connect the points to make the triangle.

🎯 Exam Tip: When the vertical angle (like \( \angle C \)) is given, construct an angle at the base (like \( \angle BAE \)) equal to \( 90^\circ \) minus the vertical angle, to help find the center of the circumcircle.

Question 16. Draw a triangle ABC of base BC = 5.6 cm, ∠A = 40° and the bisector of ∠A meets BC at D such that CD = 4 cm.
Answer:

Rough Diagram

Steps of construction:
1. Draw a line segment BC of length 5.6 cm. This is the base of the triangle.
2. At point B, draw a ray BE such that \( \angle CBE = 40^\circ \). This angle is constructed to help find the circumcenter. The angle \( \angle A \) is \( 40^\circ \), so \( \angle CBE = 40^\circ \) also.
3. At point B, draw another ray BF such that \( \angle EBF = 90^\circ \). This ray helps to locate the center of the circle.
4. Draw the perpendicular bisector of the line segment BC. This bisector will intersect the ray BF at point O and the line segment BC at point G. Point O is the center of the circumcircle.
5. Using O as the center and OB as the radius, draw a circle. The vertex A will lie on this circle.
6. From point C on BC, mark an arc of 4 cm on BC. Label this point D. This is where the angle bisector meets BC.
7. The perpendicular bisector of BC intersects the circle at point I. Join ID.
8. Extend the line segment ID until it meets the circle at point A. This determines the vertex A.
9. Join AB and AC to complete the triangle ABC. This is the required triangle.
In simple words: Start by drawing the base BC. Create a special angle at B related to angle A. Find the center of a circle that goes through A, B, and C by drawing perpendicular lines. Mark point D on BC using the given length. Draw a line from the center through D to find point A on the circle. Then, connect A to B and C to finish the triangle.

🎯 Exam Tip: When a vertical angle and an angle bisector are given, the perpendicular bisector of the base is key. Remember that the circumcenter (O) is used to draw the circle that passes through all three vertices of the triangle.

Question 17. Draw ΔPQR such that PQ = 6.8 cm, vertical angle is 50° and the bisector of the vertical angle meets the base at D where PD = 5.2 cm
Answer:

Rough Diagram

Steps of construction:
1. Draw a line segment PQ of length 6.8 cm. This forms the base of the triangle.
2. At point P, draw a ray PE such that \( \angle QPE = 50^\circ \). This angle is related to the vertical angle \( \angle R \).
3. At point P, draw another ray PF such that \( \angle EPF = 90^\circ \). This ray helps to locate the center of the circumcircle.
4. Draw the perpendicular bisector of the line segment PQ. This bisector will intersect PF at point O and PQ at point G. Point O is the center of the circumcircle.
5. Using O as the center and OP as the radius, draw a circle. The vertex R will lie on this circle.
6. From point P, mark an arc of 5.2 cm on PQ at point D. This point D is where the bisector of \( \angle R \) meets the base.
7. The perpendicular bisector intersects the circle at point I. Join ID.
8. Extend the line segment ID until it meets the circle at point A (this is referred to as 'A' in the text but should be 'R' based on the triangle PQR). Join PR and QR. The triangle PQR is the required triangle.
In simple words: Start with the base PQ. Create a special angle at P related to the vertical angle R. Find the center of a circle that includes P, Q, and R by drawing perpendicular lines. Mark point D on PQ based on the given length. Draw a line from the center through point D to find vertex R on the circle. Finally, connect the points to complete the triangle.

🎯 Exam Tip: When constructing a triangle with a given vertical angle and the length of the angle bisector to the base, correctly identifying the center of the circumcircle and using the bisector length to find the third vertex are the most critical steps for accuracy.