GSEB Class 10 Maths Solutions Chapter 6 Triangle Exercise 6.6

Question 1. In the figure, PS is the bisector of \( \angle OPR \) of \( \angle QPR \). Prove that \( = \frac { QS }{ SR } = \frac { PQ }{ PR } \)
Answer:Given: In \( \triangle PQR \), PS serves as the bisector of \( \angle QPR \).
To prove: \( \frac { QS }{ SR } = \frac { PQ }{ PR } \)
Construction: Draw a line RT parallel to PS, which meets QP produced at point T.
Proof:
\( \angle 1 = \angle 2 \) ......... (1) (This is given information)
Since RT is parallel to PS and PR is a transversal line, we get:
\( \angle 2 = \angle 3 \) (These are alternate interior angles).
Since RT is parallel to PS and QT is a transversal line, we get:
\( \angle 1 = \angle 4 \) (These are corresponding angles).
From the above equations, we can conclude that \( \angle 3 = \angle 4 \).
In \( \triangle PRT \), since \( \angle 3 = \angle 4 \), the sides opposite to these equal angles must also be equal. Therefore, PR = PT.
Now, consider \( \triangle QRT \). Since PS is parallel to RT, by the basic proportionality theorem (BPT), we have:
\( \frac { QS }{ SR } = \frac { QP }{ PT } \)
We know that QP is the same as PQ, and we just established that PT = PR. So, substituting PR for PT in the equation, we get:
\( \frac { QS }{ SR } = \frac { PQ }{ PR } \)
Thus, the desired relationship is proved.
In simple words: We are given a triangle with an angle bisector. By drawing a parallel line and using angle properties, we show that two sides become equal. Then, applying the basic proportionality theorem helps us prove the ratio of the segments is equal to the ratio of the sides.

Exam Tip: Remember to clearly state the given, to prove, and construction. Label angles and lines in your diagram for clarity, and always cite the theorems you use, like the Basic Proportionality Theorem (BPT) or properties of parallel lines and transversals.

Question 2. In the given figure, D is a point on hypotenuse AC of \( \triangle ABC \), such that \( BD \perp AC \), \( DM \perp BC \) and \( DN \perp AB \). Prove that:
(i) \( DM^2 = DN \cdot MC \)
(ii) \( DN^2 = DM \cdot AN \)
Answer:In the given figure, \( \triangle ABC \) is a right-angled triangle where \( \angle B = 90^\circ \).
We are given that \( BD \perp AC \), \( DM \perp BC \), and \( DN \perp AB \).
From the given perpendiculars, we can conclude:
\( AB \perp BC \) and \( DM \perp BC \), which implies \( AB || DM \).
Similarly, \( CB \perp AB \) and \( DN \perp AB \), which implies \( CB || DN \).
Since BMDN has opposite sides parallel and all angles are 90 degrees (due to perpendiculars), BMDN is a rectangle.
Therefore, BM = ND.
(i) To prove: \( DM^2 = DN \cdot MC \)
In \( \triangle BMD \), we have \( \angle BMD = 90^\circ \).
By the angle sum property in \( \triangle BMD \): \( \angle 1 + \angle BMD + \angle 2 = 180^\circ \).
Substituting \( \angle BMD = 90^\circ \), we get \( \angle 1 + 90^\circ + \angle 2 = 180^\circ \).
This simplifies to \( \angle 1 + \angle 2 = 90^\circ \) ...............(1)
Similarly, in \( \triangle DMC \), we have \( \angle DMC = 90^\circ \).
So, \( \angle 3 + \angle 4 = 90^\circ \) ...............(2)
Since \( BD \perp AC \), we know that \( \angle BDC = 90^\circ \).
In \( \triangle BDC \), we have \( \angle BDC = 90^\circ \), so \( \angle 2 + \angle 3 = 90^\circ \) ...............(3)
From equations (1) and (3), we can equate the sums:
\( \angle 1 + \angle 2 = \angle 2 + \angle 3 \)
Subtracting \( \angle 2 \) from both sides, we get \( \angle 1 = \angle 3 \).
From equations (2) and (3), we can also equate:
\( \angle 3 + \angle 4 = \angle 2 + \angle 3 \)
Subtracting \( \angle 3 \) from both sides, we get \( \angle 2 = \angle 4 \).
Now, consider \( \triangle DBN \) and \( \triangle DCM \).
We have \( \angle BND = \angle DMC = 90^\circ \).
Also, \( \angle DBN = \angle DCM \) (since \( \angle 1 = \angle 3 \)).
Therefore, \( \triangle DBN \sim \triangle DCM \) (by AA similarity criterion).
When triangles are similar, the ratios of their corresponding sides are equal:
\( \frac { DN }{ DM } = \frac { BN }{ CM } = \frac { DB }{ DC } \)
From \( \frac { DN }{ DM } = \frac { BN }{ CM } \), we get \( DN \cdot CM = DM \cdot BN \).
We know that in rectangle BMDN, BN = DM. So, substituting DM for BN:
\( DN \cdot CM = DM \cdot DM \)
Thus, \( DM^2 = DN \cdot MC \). (This completes part i. The OCR seems to have reversed the parts of question ii and the final result).
(ii) To prove: \( DN^2 = DM \cdot AN \)
Consider \( \triangle ADN \) and \( \triangle BDM \).
We know \( \angle DNA = \angle DMB = 90^\circ \).
From earlier, \( \angle 2 = \angle 4 \).
Therefore, \( \triangle ADN \sim \triangle BDM \) (by AA similarity criterion).
When triangles are similar, the ratios of their corresponding sides are equal:
\( \frac { DN }{ DM } = \frac { AN }{ BM } = \frac { AD }{ BD } \)
From \( \frac { DN }{ DM } = \frac { AN }{ BM } \), we get \( DN \cdot BM = DM \cdot AN \).
We know that in rectangle BMDN, BM = DN. So, substituting DN for BM:
\( DN \cdot DN = DM \cdot AN \)
Thus, \( DN^2 = DM \cdot AN \). (This completes part ii).
In simple words: First, we use the fact that the given figure BMDN is a rectangle because of all the perpendicular lines, which means its opposite sides are equal. Then, by identifying similar triangles using angle properties, we can set up ratios of their sides. Cross-multiplying these ratios and substituting the equal sides from the rectangle helps us prove both required equations.

Exam Tip: For problems involving multiple perpendiculars and similar triangles, always begin by identifying any special quadrilaterals (like rectangles or squares) formed. This will give you important side equalities which are very useful in later similarity proofs.

Question 3. In figure ABC is a triangle in which \( \angle ABC > 90^\circ \) and \( AD \perp CB \) produced prove that \( AC^2 = AB^2 + BC^2 + 2BC \cdot BD \)
Answer:Given: \( \triangle ABC \) where \( \angle ABC > 90^\circ \). AD is perpendicular to CB produced, meaning \( AD \perp CB \) (produced).
To prove: \( AC^2 = AB^2 + BC^2 + 2BC \cdot BD \)
Proof:
Consider the right-angled triangle \( \triangle ADC \). Here, \( \angle D = 90^\circ \).
By applying the Pythagoras theorem in \( \triangle ADC \), we have:
\( AC^2 = AD^2 + DC^2 \)
We can express DC as the sum of DB and BC: \( DC = BD + BC \).
Substitute this into the equation:
\( AC^2 = AD^2 + (BD + BC)^2 \)
Expand the term \( (BD + BC)^2 \):
\( AC^2 = AD^2 + BD^2 + BC^2 + 2BD \cdot BC \) ...............(1)
Now, consider the right-angled triangle \( \triangle ADB \). Here, \( \angle D = 90^\circ \).
By applying the Pythagoras theorem in \( \triangle ADB \), we have:
\( AB^2 = AD^2 + BD^2 \) ...............(2)
Now, substitute the expression for \( (AD^2 + BD^2) \) from equation (2) into equation (1):
\( AC^2 = AB^2 + BC^2 + 2BC \cdot BD \)
Thus, the relationship is proved.
In simple words: For an obtuse triangle, if we drop a perpendicular from the vertex opposite the obtuse angle to the extended base, we can use Pythagoras' theorem twice. First on the larger right-angled triangle, and then on the smaller one formed by the altitude. By substituting one equation into the other, we get the desired relationship.

Exam Tip: This is a generalization of the Pythagorean theorem for obtuse triangles. Always identify the right-angled triangles created by the altitude and use the segment addition/subtraction property for the base correctly. Double-check your algebraic expansion of squared binomials.

Question 4. In figure, ABC is a triangle in which \( \angle ABC < 90^\circ \) and \( AD \perp BC \). To Prove: \( AC^2 = AB^2 + BC^2 - 2BC \cdot BD \)
Answer:Given: \( \triangle ABC \) where \( \angle ABC < 90^\circ \). AD is perpendicular to BC, meaning \( AD \perp BC \).
To prove: \( AC^2 = AB^2 + BC^2 - 2BC \cdot BD \)
Proof:
Consider the right-angled triangle \( \triangle ADC \). Here, \( \angle D = 90^\circ \).
By applying the Pythagoras theorem in \( \triangle ADC \), we have:
\( AC^2 = AD^2 + DC^2 \)
We can express DC as the difference between BC and BD: \( DC = BC - BD \).
Substitute this into the equation:
\( AC^2 = AD^2 + (BC - BD)^2 \)
Expand the term \( (BC - BD)^2 \):
\( AC^2 = AD^2 + BC^2 + BD^2 - 2BC \cdot BD \) ...............(1)
Now, consider the right-angled triangle \( \triangle ADB \). Here, \( \angle D = 90^\circ \).
By applying the Pythagoras theorem in \( \triangle ADB \), we have:
\( AB^2 = AD^2 + BD^2 \) ...............(2)
Now, substitute the expression for \( (AD^2 + BD^2) \) from equation (2) into equation (1):
\( AC^2 = AB^2 + BC^2 - 2BC \cdot BD \)
Thus, the relationship is proved.
In simple words: This is an extension of Pythagoras' theorem for acute triangles. We draw an altitude, forming two right-angled triangles. By applying Pythagoras' theorem to both and using the fact that one segment of the base is the total base minus another segment, we can derive the formula.

Exam Tip: Pay close attention to whether the angle is acute or obtuse, as this determines the sign of the \(2BC \cdot BD\) term. For acute angles, the perpendicular falls inside the triangle, leading to subtraction; for obtuse, it falls outside, leading to addition (as seen in Question 3).

Question 5. In figure AD is a median of a triangle ABC and AM \( \perp \) BC prove that
(i) \( AC^2 = AD^2 + BC \cdot DM + \left(\frac { BC }{ 2 } \right)^2 \)
(ii) \( AB^2 = AD^2 - BC \cdot DM + \left(\frac { BC }{ 2 } \right)^2 \)
(iii) \( AC^2 + AB^2 = 2AD^2 + \frac { 1 }{ 2 } BC^2 \)
Answer:Given: AD serves as a median of \( \triangle ABC \), which means D is the midpoint of BC. AM is perpendicular to BC, so \( AM \perp BC \).
Proof:
(i) To prove: \( AC^2 = AD^2 + BC \cdot DM + \left(\frac { BC }{ 2 } \right)^2 \)
Consider the right-angled triangle \( \triangle AMC \). Here, \( \angle M = 90^\circ \).
By the Pythagoras theorem, \( AC^2 = AM^2 + MC^2 \).
Since D is the midpoint of BC, \( DC = \frac{BC}{2} \).
We can write MC as \( MC = DM + DC \).
Substitute this into the equation for \( AC^2 \):
\( AC^2 = AM^2 + (DM + DC)^2 \)
Expand the term \( (DM + DC)^2 \):
\( AC^2 = AM^2 + DM^2 + DC^2 + 2DM \cdot DC \)
Now, substitute \( DC = \frac{BC}{2} \):
\( AC^2 = AM^2 + DM^2 + \left(\frac { BC }{ 2 } \right)^2 + 2DM \cdot \left(\frac { BC }{ 2 } \right) \)
\( AC^2 = AM^2 + DM^2 + \left(\frac { BC }{ 2 } \right)^2 + DM \cdot BC \) ...............(1)
Next, consider the right-angled triangle \( \triangle AMD \). Here, \( \angle M = 90^\circ \).
By the Pythagoras theorem, \( AD^2 = AM^2 + DM^2 \) ...............(2)
Substitute \( AD^2 \) from equation (2) into equation (1):
\( AC^2 = AD^2 + BC \cdot DM + \left(\frac { BC }{ 2 } \right)^2 \). (Part i is proved)
(ii) To prove: \( AB^2 = AD^2 - BC \cdot DM + \left(\frac { BC }{ 2 } \right)^2 \)
Consider the right-angled triangle \( \triangle AMB \). Here, \( \angle M = 90^\circ \).
By the Pythagoras theorem, \( AB^2 = AM^2 + BM^2 \).
Since D is the midpoint of BC, \( BD = \frac{BC}{2} \).
We can write BM as \( BM = BD - DM \).
Substitute this into the equation for \( AB^2 \):
\( AB^2 = AM^2 + (BD - DM)^2 \)
Expand the term \( (BD - DM)^2 \):
\( AB^2 = AM^2 + BD^2 + DM^2 - 2BD \cdot DM \)
Now, substitute \( BD = \frac{BC}{2} \):
\( AB^2 = AM^2 + DM^2 + \left(\frac { BC }{ 2 } \right)^2 - 2 \left(\frac { BC }{ 2 } \right) \cdot DM \)
\( AB^2 = AM^2 + DM^2 + \left(\frac { BC }{ 2 } \right)^2 - BC \cdot DM \) ...............(3)
Substitute \( AD^2 \) from equation (2) into equation (3):
\( AB^2 = AD^2 - BC \cdot DM + \left(\frac { BC }{ 2 } \right)^2 \). (Part ii is proved)
(iii) To prove: \( AC^2 + AB^2 = 2AD^2 + \frac { 1 }{ 2 } BC^2 \)
Add the equations from part (i) and part (ii):
\( AC^2 + AB^2 = \left(AD^2 + BC \cdot DM + \left(\frac { BC }{ 2 } \right)^2\right) + \left(AD^2 - BC \cdot DM + \left(\frac { BC }{ 2 } \right)^2\right) \)
Combine like terms:
\( AC^2 + AB^2 = AD^2 + AD^2 + BC \cdot DM - BC \cdot DM + \left(\frac { BC }{ 2 } \right)^2 + \left(\frac { BC }{ 2 } \right)^2 \)
The \( BC \cdot DM \) terms cancel out:
\( AC^2 + AB^2 = 2AD^2 + 2 \left(\frac { BC }{ 2 } \right)^2 \)
Simplify the squared term:
\( AC^2 + AB^2 = 2AD^2 + 2 \left(\frac { BC^2 }{ 4 } \right) \)
\( AC^2 + AB^2 = 2AD^2 + \frac { BC^2 }{ 2 } \). (Part iii is proved)
In simple words: This problem uses Apollonius' theorem, which relates the sides of a triangle to the length of a median. We achieve this by applying the extended Pythagorean theorem (from Q3 and Q4) to the two triangles formed by the median and altitude, then adding or subtracting the resulting equations to derive the required identities.

Exam Tip: This problem demonstrates Apollonius' theorem and its proof. Remember that a median divides the opposite side into two equal parts. Carefully handle the algebraic expansions and substitutions, especially when terms cancel out. Understanding the general forms for acute and obtuse triangles is crucial here.

Question 6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.
Answer:Given: ABCD is a parallelogram whose diagonals are AC and BD.
To prove: \( AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2 \)
Construction: Draw AM perpendicular to DC and BN perpendicular to DC (produced).
Proof:
In parallelogram ABCD, we know that opposite sides are equal: AD = BC and AB = CD.
Also, AM and BN are altitudes drawn to the same base (or its extension), so AM = BN.
Consider \( \triangle AMD \) and \( \triangle BNC \).
We have AD = BC (opposite sides of a parallelogram).
\( \angle AMD = \angle BNC = 90^\circ \) (by construction).
\( AM = BN \) (altitudes on the same base).
Therefore, \( \triangle AMD \cong \triangle BNC \) (by RHS congruence rule).
By CPCT (Corresponding Parts of Congruent Triangles), we have MD = NC ...............(1)
Now, consider \( \triangle BND \), which is a right-angled triangle with \( \angle N = 90^\circ \).
By the Pythagoras theorem:
\( BD^2 = BN^2 + DN^2 \)
We can write DN as \( DN = DC + CN \).
So, \( BD^2 = BN^2 + (DC + CN)^2 \)
Expand the term \( (DC + CN)^2 \):
\( BD^2 = BN^2 + DC^2 + CN^2 + 2DC \cdot CN \) ...............(2)
Also, consider \( \triangle BNC \), which is a right-angled triangle.
By the Pythagoras theorem: \( BC^2 = BN^2 + CN^2 \) ...............(3)
Substitute \( BC^2 \) from equation (3) into equation (2):
\( BD^2 = BC^2 + DC^2 + 2DC \cdot CN \) ...............(4)
Next, consider \( \triangle AMC \), which is a right-angled triangle with \( \angle M = 90^\circ \).
By the Pythagoras theorem:
\( AC^2 = AM^2 + MC^2 \)
We can write MC as \( MC = DC - DM \).
So, \( AC^2 = AM^2 + (DC - DM)^2 \)
Expand the term \( (DC - DM)^2 \):
\( AC^2 = AM^2 + DC^2 + DM^2 - 2DC \cdot DM \) ...............(5)
Also, consider \( \triangle AMD \), which is a right-angled triangle.
By the Pythagoras theorem: \( AD^2 = AM^2 + DM^2 \) ...............(6)
Substitute \( AD^2 \) from equation (6) into equation (5):
\( AC^2 = AD^2 + DC^2 - 2DC \cdot DM \) ...............(7)
We know that in a parallelogram, opposite sides are equal (DC = AB) and from equation (1), DM = CN.
Now, add equation (4) and equation (7):
\( BD^2 + AC^2 = (BC^2 + DC^2 + 2DC \cdot CN) + (AD^2 + DC^2 - 2DC \cdot DM) \)
Substitute DM with CN (since they are equal from equation 1) and simplify:
\( BD^2 + AC^2 = BC^2 + DC^2 + 2DC \cdot CN + AD^2 + DC^2 - 2DC \cdot CN \)
The terms \( 2DC \cdot CN \) and \( -2DC \cdot CN \) cancel each other out.
\( AC^2 + BD^2 = BC^2 + DC^2 + AD^2 + DC^2 \)
Since DC = AB (opposite sides of a parallelogram), substitute AB for one of the DC terms:
\( AC^2 + BD^2 = AB^2 + BC^2 + CD^2 + DA^2 \)
Thus, the sum of the squares of the diagonals of a parallelogram equals the sum of the squares of its sides.
In simple words: To prove this property of parallelograms, we draw altitudes from the vertices to the base. By using the congruence of certain triangles and applying Pythagoras' theorem to the right-angled triangles formed, we derive expressions for the diagonals. Adding these expressions leads to the desired result, showing the relationship between diagonals and sides.

Exam Tip: This proof is a direct application of the Apollonius theorem twice (once for each diagonal) and then adding them. Ensure you correctly identify congruent triangles and apply the Pythagorean theorem. Remember to use the properties of parallelograms, such as opposite sides being equal, to simplify your equations.

Question 7. In figure, two chords AB and CD intersect each other at the point P prove that
(i) \( \triangle APC \sim \triangle DPB \)
(ii) \( AP \cdot PB = CP \cdot DP \)
Answer:Given: Two chords AB and CD of a circle intersect each other at point P.
To prove:
(i) \( \triangle APC \sim \triangle DPB \)
(ii) \( AP \cdot PB = CP \cdot DP \)
Proof:
(i) Consider \( \triangle APC \) and \( \triangle DPB \).
We have \( \angle APC = \angle DPB \) (These are vertically opposite angles, and vertically opposite angles are always equal).
We also have \( \angle CAP = \angle BDP \) (These are angles subtended by the same arc, arc CB, in the same segment of the circle. Angles in the same segment are equal).
Therefore, by the Angle-Angle (AA) similarity criterion, we can conclude:
\( \triangle APC \sim \triangle DPB \). (Part i is proved)
(ii) Since \( \triangle APC \sim \triangle DPB \) (as proved above), the ratios of their corresponding sides must be proportional.
So, \( \frac { AP }{ DP } = \frac { CP }{ PB } = \frac { AC }{ DB } \)
Taking the first two parts of the proportion:
\( \frac { AP }{ DP } = \frac { CP }{ PB } \)
Cross-multiplying these terms, we get:
\( AP \cdot PB = CP \cdot DP \). (Part ii is proved)
In simple words: When two chords cross inside a circle, they form two pairs of similar triangles. We prove this similarity by showing that their vertically opposite angles are equal, and their angles subtended by the same arc are also equal. Once similar, the ratios of their corresponding sides are equal, which lets us prove the product rule for intersecting chords.

Exam Tip: This theorem is about intersecting chords. The key steps are recognizing vertically opposite angles and angles in the same segment, which together prove triangle similarity by AA. Once similarity is established, the proportionality of sides directly leads to the product rule for intersecting chords.

Question 8. In figure, two chords AB and CD of a circle intersect each other at the point P (When produce) outside the circle. Prove that
(i) \( \triangle PAC \sim \triangle PDB \)
(ii) \( PA \cdot PB = PC \cdot PD \)
Answer:Given: Two chords AB and CD of a circle are produced to intersect each other at point P outside the circle.
To prove:
(i) \( \triangle PAC \sim \triangle PDB \)
(ii) \( PA \cdot PB = PC \cdot PD \)
Proof:
(i) Consider \( \triangle PAC \) and \( \triangle PDB \).
Both triangles share a common angle, \( \angle P = \angle P \).
In a cyclic quadrilateral (like ABCD formed by the chords), the exterior angle equals the interior opposite angle.
Therefore, \( \angle PAC = \angle PDB \) (The angle \( \angle PAC \) is an exterior angle to the cyclic quadrilateral ABCD at vertex A, which is equal to the interior opposite angle \( \angle D \), i.e., \( \angle PDB \)).
Also, \( \angle PCA = \angle PBD \) (Similarly, \( \angle PCA \) is an exterior angle to the cyclic quadrilateral ABCD at vertex C, which is equal to the interior opposite angle \( \angle B \), i.e., \( \angle PBD \)).
Therefore, by the Angle-Angle (AA) similarity criterion, we can conclude:
\( \triangle PAC \sim \triangle PDB \). (Part i is proved)
(ii) Since \( \triangle PAC \sim \triangle PDB \) (as proved above), the ratios of their corresponding sides must be proportional.
So, \( \frac { PA }{ PD } = \frac { PC }{ PB } = \frac { AC }{ DB } \)
Taking the first two parts of the proportion:
\( \frac { PA }{ PD } = \frac { PC }{ PB } \)
Cross-multiplying these terms, we get:
\( PA \cdot PB = PC \cdot PD \). (Part ii is proved)
In simple words: When chords of a circle extend and meet outside, they form similar triangles. We prove similarity by showing they share a common angle and that the exterior angle of the cyclic quadrilateral formed by the chords is equal to the interior opposite angle. This similarity then directly leads to the product rule for external intersecting chords.

Exam Tip: This theorem is about external intersecting chords. The key is to recognize the common angle at P and use the property of cyclic quadrilaterals (exterior angle equals interior opposite angle) to establish similarity. Once similarity is proven by AA, the proportionality of sides yields the product rule for external segments.

Question 9. In figure, D is a point on side BC of \( \triangle ABC \) such that \( \frac { BD }{ CD } = \frac { AB }{ AC } \). Prove that AD is the bisector of \( \angle BAC \).
Answer:Given: In \( \triangle ABC \), D is a point on side BC such that \( \frac { BD }{ CD } = \frac { AB }{ AC } \).
To prove: AD is the bisector of \( \angle BAC \), which means \( \angle BAD = \angle CAD \).
Construction: Produce BA to a point E such that AE = AC. Then, join C to E.
Proof:
We are given \( \frac { BD }{ CD } = \frac { AB }{ AC } \).
From our construction, we know that AE = AC. So, we can substitute AE for AC in the given ratio:
\( \frac { BD }{ CD } = \frac { AB }{ AE } \)
Now, in \( \triangle BCE \), we have point D on BC and point A on BE. From the above ratio, by the converse of the Basic Proportionality Theorem (BPT), if a line divides two sides of a triangle proportionally, then it is parallel to the third side. Therefore:
AD \( || \) CE ...............(1)
Since AD is parallel to CE, and AC is a transversal line, we can say:
\( \angle 2 = \angle 4 \) (These are alternate interior angles). ...............(2)
Also, since AD is parallel to CE, and BE is a transversal line, we can say:
\( \angle 1 = \angle 3 \) (These are corresponding angles). ...............(3)
In \( \triangle ACE \), we constructed AE = AC. When two sides of a triangle are equal, the angles opposite those sides are also equal. Therefore:
\( \angle 3 = \angle 4 \).
Now, combining equations (2) and (3), since \( \angle 1 = \angle 3 \) and \( \angle 2 = \angle 4 \), and we know \( \angle 3 = \angle 4 \), it follows that:
\( \angle 1 = \angle 2 \).
Since \( \angle BAD = \angle CAD \) (which are \( \angle 1 \) and \( \angle 2 \) respectively), we have proved that AD is the bisector of \( \angle BAC \).
In simple words: This problem proves the Angle Bisector Theorem. We start with the given ratio of side segments and use a construction to create a parallel line. Then, by applying properties of parallel lines (alternate interior and corresponding angles) and the property of an isosceles triangle, we show that the two angles formed by the line AD are equal, proving it's an angle bisector.

Exam Tip: This is the converse of the Angle Bisector Theorem. The crucial step is the construction of a parallel line (CE) and using the property of isosceles triangles (AE = AC implies \( \angle 3 = \angle 4 \)). Clearly identifying alternate interior and corresponding angles due to parallel lines is also essential.

Question 10. A girl is fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?
Answer:First, let's determine the initial length of the string the girl has out.
Consider the right-angled triangle formed by the tip of the rod (A), the point directly under it on the water (B), and the fly on the water (C). The question states the fly is 2.4m from the point directly under the rod, and the tip of the rod is 1.8m above the water.
So, we have a right-angled triangle with sides AB (height of rod tip) = 1.8 m and BC (horizontal distance from under tip to fly) = 2.4 m.
Let AC be the length of the string. By the Pythagoras theorem:
\( AC^2 = AB^2 + BC^2 \)
\( AC^2 = (1.8)^2 + (2.4)^2 \)
\( AC^2 = 3.24 + 5.76 \)
\( AC^2 = 9 \)
\( AC = \sqrt{9} \)
\( AC = 3 \) m
Therefore, the length of the string she has out initially is 3 meters.
Now, for the second part of the question: If she pulls in the string at a rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?
Rate of pulling in the string = 5 cm/second.
Amount of string pulled in during 12 seconds = \( 5 \text{ cm/s} \times 12 \text{ s} = 60 \text{ cm} \).
Convert 60 cm to meters: \( \frac { 60 }{ 100 } \text{ m} = 0.60 \text{ meter} \).
Remaining string length after 12 seconds = (Initial length) - (Pulled in length)
Remaining string length = \( 3 \text{ m} - 0.60 \text{ m} = 2.4 \text{ m} \).
Let this new string length be \( AC' = 2.4 \text{ m} \). The height of the rod tip (AB) remains 1.8 m.
We want to find the new horizontal distance, let's call it \( BC' \).
Again, using the Pythagoras theorem for the new right-angled triangle (A, B, C'):
\( (AC')^2 = AB^2 + (BC')^2 \)
\( (2.4)^2 = (1.8)^2 + (BC')^2 \)
\( 5.76 = 3.24 + (BC')^2 \)
\( (BC')^2 = 5.76 - 3.24 \)
\( (BC')^2 = 2.52 \)
\( BC' = \sqrt{2.52} \)
\( BC' \approx 1.59 \) m (approximately)
This is the new horizontal distance of the fly from the point directly under the tip of the rod.
The question asks for the horizontal distance of the fly *from her* (assuming "her" refers to the point where the rod is held, which is above the point directly under the tip). The problem states the fly is initially 3.6m away, and 2.4m from the point *directly under the tip*. This implies the rod is held 1.2m behind the point directly under the tip, as shown in the diagram (2.4m + 1.2m = 3.6m).
So, the horizontal distance from her to the fly after 12 seconds will be the new horizontal distance from the point under the tip plus the 1.2m distance she is behind that point.
Total horizontal distance from her = \( 1.59 \text{ m} + 1.2 \text{ m} = 2.79 \text{ m} \) (approximately).
In simple words: First, we use the Pythagorean theorem to find the length of the fishing line. Then, we calculate how much line is reeled in over 12 seconds. Subtracting this from the total line gives us the new line length. Using Pythagoras again, we find the new horizontal distance from the rod's tip. Finally, we add the fixed horizontal distance from her position to the rod's tip to get the total horizontal distance.

Exam Tip: Break down word problems into smaller steps. Clearly identify the right-angled triangles and known values for each step. Pay close attention to unit conversions (cm to m) and whether the question asks for distance from the rod tip or from the person holding the rod, as extra distances might need to be added or subtracted.