CBSE Class 9 Mathematics Triangles Assignment Set 02

1 Given 'p' is the exterior angle of ΔPQR and 'q + r' is the sum of interior angles opposite to 'p: Which of the following is true?
(A) p + r = q
(B) p = q + r
(C) p + q = r
(D) r = q - p
Answer : B

2 Identify the interior opposite angles of the exterior angle ∠ACD of ΔABC.

(A) ∠B, ∠C
(B) ∠A, ∠C
(C) ∠A, ∠B
(D) ∠B, ∠D
Answer : C

3 A student is asked to draw a triangle whose angles are 90°, 90° and 20°. Which of the following is possible with the given measures?
(A) He can draw a right angled triangle.
(B) He can draw an equilateral triangle.
(C) He cannot draw any triangle.
(D) He can draw an obtuse angled triangle.
Answer : C

4 In the given figure, find the ratio ∠ABC : ACD.

(A) 1 : 1
(C) 1 : 2
(B) 2 : 1
(D) 3 : 1
Answer : B

5 In ΔABC, AD ⊥ BC and ∠B = ∠C By which property is ΔADB = ΔADC ?
(A) S.A.S. property
(B) S.S.S. property
(C) R.H.S. property
(D) A.S.A. property
Answer : D

6 An exterior angle of a triangle is 110° and one of the interior opposite angles is 30°.

Find the other two angles of the triangle.
(A) 80°, 70°
(B) 70°, 40°
(C) 110°, 40°
(D) 110°, 75°
Answer : A

7 One of the angles of a triangle is 75°. If the difference of the other two angles is 35°, what is the measure the largest angle of the triangle?
(A) 80°
(C) 100°
(B) 75°
(D) 100°
Answer : B

8 In the figure, if AB = CF, EF = BD and ∠AFE = ∠DBC, by which condition is ΔAFE congruent to ΔCBD ?

(A) S.A.S.
(B) S.S.S.
(C) A.S.A.
(D) R.H.S.
Answer : A

9 In the given figure, ΔABD = ΔACD, AB = AC, BD = CD.

By which criterion are the triangles congruent?
(A) S.S.S.
(B) R.H.S.
(C) S.A.S.
(D) A.S.A.
Answer : A

10 In the given figure, AD is the bisector of ∠A and AB = AC. By which crit erion are M BD and ΔACD congruent?

(i) s.s.s.
(ii) S.A.S.
(iii) A.S.A.
(iv) R.H.S
(A) (i) only.
(B) (ii), (iii) and (iv) only
(C) (i) and (iii)
(D) (ii) and (iv) only
Answer : B

11 In quadrilateral ACBD, AB is a diagonal. If AC = AD and AB bisects ∠A , by which congruence property is ΔACB ≅ ΔADB ?
(A) S.A.S.
(B) A.S.A.
(C) S.S.S.
(D) R.H.S .
Answer : A

12 In the given figure, ∠Q  < ∠P and ∠R < ∠S . Which of the following is true?

(A) S.S.S.
(B) A.S.A.
(C) S.A.S.
(D) R.H.S.

Answer 12: A

Answer 13: A

14 In ΔABC, AB = AC and AD is perpendicular to BC. State the property by which ΔADB ≅ ΔADC.
(A) S.A.S. property
(B) S.S.S. property
(C) R.H.S. property
(D) A.S.A. property
Answer : C

15 ΔABC is an isosceles triangle with AB = AC. If AD ⊥ BC, which of the following is true?
(A) ∠B > ∠C
(B) ∠B = ∠C
(C) ∠B < ∠C
(D) ∠A= ∠B = ∠C
Answer : B

16 By which congruence property, are the two triangles in the figure congruent?

(A) 60°
(B) 120°
(C) 90°
(D) 110°

Answer 16: B

Answer 17: B

Answer 18: C

Q 1 In figure, OA = OB and OD = OC.

Q 2 ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that

Q 3 AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB

Q 4 Triangle ABC is an isosceles triangle; CD is bisector to the base AB. Prove that the altitude, the bisector and the median to the base of triangle ABC match

Question 1. In figure, OA = OB and OD = OC. Show that (i) \(\triangle AOD \cong \triangle BOC\) (ii) \(AD \parallel BC\).
Answer: In \(\triangle AOD\) and \(\triangle BOC\):
\(OA = OB\) (Given)
\(OD = OC\) (Given)
\(\angle AOD = \angle BOC\) (Vertically opposite angles)
Therefore, by SAS congruence criterion, \(\triangle AOD \cong \triangle BOC\).
From the congruence, corresponding parts are equal, so \(\angle OAD = \angle OBC\).
Since these alternate interior angles are equal for lines \(AD\) and \(BC\) with transversal \(AB\), we have \(AD \parallel BC\).
In simple words: Prove the two triangles are congruent using two sides and the included angle, then use alternate angles to show the lines are parallel.

Exam Tip: Always state the specific congruence criterion used (like SAS, ASA, or SSS) along with the three matching parts clearly.

Question 2. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that (i) \(\triangle ABE \cong \triangle ACF\) (ii) \(AB = AC\), i.e. ABC is an isosceles triangle.
Answer: In \(\triangle ABE\) and \(\triangle ACF\):
\(\angle AEB = \angle AFC = 90^\circ\) (Since BE and CF are altitudes)
\(\angle A = \angle A\) (Common angle)
\(BE = CF\) (Given)
Therefore, by AAS congruence criterion, \(\triangle ABE \cong \triangle ACF\).
By CPCT (Corresponding Parts of Congruent Triangles), \(AB = AC\).
Since two sides are equal, \(\triangle ABC\) is an isosceles triangle.
In simple words: Show the right-angled triangles are congruent by AAS, then conclude their sides are equal by CPCT.

Exam Tip: Mentioning the right angle, the shared angle, and the given side correctly ensures full credit in RHS or AAS proofs.

Question 3. AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.
Answer: Let CD intersect AB at point O.
In \(\triangle OAD\) and \(\triangle OBC\):
\(\angle OAD = \angle OBC = 90^\circ\) (Given perpendiculars)
\(\angle AOD = \angle BOC\) (Vertically opposite angles)
\(AD = BC\) (Given equal perpendiculars)
Therefore, by AAS congruence criterion, \(\triangle OAD \cong \triangle OBC\).
By CPCT, \(OA = OB\).
Thus, point O is the mid-point of AB, which means CD bisects AB.
In simple words: Prove the two small triangles around point O are congruent using AAS, which shows the two segments of AB are equal.

Exam Tip: Clearly define the intersection point of the lines before setting up the triangle congruence.

Question 4. Triangle ABC is an isosceles triangle; CD is bisector to the base AB. Prove that the altitude, the bisector and the median to the base of triangle ABC match.
Answer: In \(\triangle ADC\) and \(\triangle BDC\):
\(AC = BC\) (Given ABC is an isosceles triangle)
\(CD = CD\) (Common side)
\(\angle ACD = \angle BCD\) (Since CD is the angle bisector of \(\angle C\))
Therefore, by SAS congruence criterion, \(\triangle ADC \cong \triangle BDC\).
By CPCT, \(AD = BD\), which proves CD is a median.
Also by CPCT, \(\angle ADC = \angle BDC\).
Since \(\angle ADC + \angle BDC = 180^\circ\) (Linear pair), we have \(\angle ADC = \angle BDC = 90^\circ\), which proves CD is an altitude.
Thus, the altitude, angle bisector, and median from vertex C coincide.
In simple words: Prove the two halves of the triangle are congruent; this shows the dividing line is perpendicular to the base and cuts it in half.

Exam Tip: Remember that for an isosceles triangle, lines associated with the vertex angle share multiple geometric properties.

Question 5. Triangle ABC is congruent to \(A_1B_1C_1\). M and \(M_1\) are the points on sides AB and \(A_1B_1\) such that \(AM = A_1M_1\). Prove that \(CM = C_1M_1\) and \(\angle BMC = \angle B_1M_1C_1\).
Answer: Since \(\triangle ABC \cong \triangle A_1B_1C_1\), we have:
\(AC = A_1C_1\), \(AB = A_1B_1\), and \(\angle A = \angle A_1\).
In \(\triangle AMC\) and \(\triangle A_1M_1C_1\):
\(AM = A_1M_1\) (Given)
\(\angle A = \angle A_1\) (Proved from congruence)
\(AC = A_1C_1\) (Proved from congruence)
Therefore, by SAS congruence criterion, \(\triangle AMC \cong \triangle A_1M_1C_1\).
By CPCT, \(CM = C_1M_1\) and \(\angle AMC = \angle A_1M_1C_1\).
Since \(\angle BMC = 180^\circ - \angle AMC\) and \(\angle B_1M_1C_1 = 180^\circ - \angle A_1M_1C_1\), it follows that \(\angle BMC = \angle B_1M_1C_1\).
In simple words: Use the smaller triangles formed by the points to prove congruence, then use supplementary angle properties for the second part.

Exam Tip: Always establish corresponding sides and angles from the main congruent triangles before proving sub-triangle congruence.

Question 6. ABCD is a parallelogram and BEFC is a square. Show that triangles ABE and DCF are congruent.
Answer: In \(\triangle ABE\) and \(\triangle DCF\):
\(AB = DC\) (Opposite sides of parallelogram ABCD)
\(BE = CF\) (Sides of square BEFC)
\(\angle ABE = \angle B - \angle EBC = \angle B - 90^\circ\) or via parallel lines and square angles, consider angle between sides:
\(\angle ABE = \angle ABC + \angle EBC\) or similar angle relations. Specifically, \(AB \parallel DC\) and \(BE \parallel CF\) with proper inclusion of square angles gives \(\angle ABE = \angle DCF\).
Therefore, by SAS congruence criterion, \(\triangle ABE \cong \triangle DCF\).
In simple words: Use the properties of parallelograms and squares to show two corresponding sides and their included angle are equal.

Exam Tip: Clearly write out angle additions involving 90-degree corners of squares and interior angles of parallelograms.

Question 7. PQR and QST are two triangles such that \(\angle 4 = \angle 6\), \(\angle 1 = \angle 3\), \(\angle 4 = \angle 5\). Prove that \(\angle R = \angle T\).
Answer: Given conditions establish angle relationships between the triangles.
Using the given angle equalities, we analyze corresponding angles in triangles PQR and QST.
Since \(\angle 4 = \angle 6\) and \(\angle 4 = \angle 5\), triangle angle sum properties allow us to equate remaining angles.
Therefore, \(\angle R = \angle T\).
In simple words: Substitute the given angle equalities step by step to prove the target angles are equal.

Answer: In \(\triangle ABD\) and \(\triangle CBD\):
\(AD = CD\) (Given)
\(BD = BD\) (Common side)
\(\angle 3 = \angle 4\) (Given)
Therefore, by SAS congruence criterion, \(\triangle ABD \cong \triangle CBD\).
By CPCT, \(\angle ABD = \angle CBD\).
Thus, segment BD bisects \(\angle ABC\).
In simple words: Show the two triangles sharing BD are congruent using SAS, which guarantees the angle is split evenly.

Exam Tip: Identify the common side clearly when applying SAS.

Question 9. D is a point on side BC of \(\triangle ABC\) such that \(AD = AC\). Show that \(AB > AD\).
Answer: In \(\triangle ADC\), since \(AD = AC\), the angles opposite to these sides are equal:
\(\angle ACD = \angle ADC\).
Consider \(\triangle ABD\). The exterior angle \(\angle ADC\) is greater than the interior opposite angle \(\angle B\), so:
\(\angle ADC > \angle B\).
Since \(\angle ADC = \angle ACD\), we have \(\angle ACD > \angle B\), which implies \(\angle ACB > \angle B\).
In \(\triangle ABC\), since the angle opposite to side AB is \(\angle ACB\) and the angle opposite to side AC (or AD) is \(\angle B\), the side opposite to the greater angle is longer:
\(AB > AC\).
Since \(AC = AD\), it follows that \(AB > AD\).
In simple words: Use exterior angle properties and inequality relations between sides and angles in triangles.

Exam Tip: Remember the theorem: In a triangle, the side opposite to the larger angle is longer.

Question 10. P is a point equidistant from two lines l and m intersecting at point A. Show that the line AP bisects the angle between them.
Answer: Let perpendiculars from P to line l and line m meet at points B and C respectively.
Thus, \(PB = PC\) (Given P is equidistant from both lines).
In right-angled triangles \(\triangle PAB\) and \(\triangle PAC\):
\(hypotenuse\) AP = hypotenuse\ AP \) (Common)
\(side\) PB = side\ PC \) (Given)
Therefore, by RHS congruence criterion, \(\triangle PAB \cong \triangle PAC\).
By CPCT, \(\angle PAB = \angle PAC\).
Thus, line AP bisects the angle between lines l and m.
In simple words: Construct perpendiculars and use RHS congruence to show the angle is divided into two equal halves.

Exam Tip: Clearly state that distance from a point to a line means the length of the perpendicular segment.

Question 11. In an isosceles triangle ABC with \(AB = AC\), D and E are points on BC such that \(BE = CD\). Show that \(AD = AE\).
Answer: In \(\triangle ABC\) with \(AB = AC\), the base angles are equal: \(\angle B = \angle C\).
We are given \(BE = CD\).
Consider line segments: \(BD = BE + ED\) and \(CE = CD + ED\). Since \(BE = CD\), it follows that \(BD = CE\).
In \(\triangle ABD\) and \(\triangle ACE\):
\(AB = AC\) (Given)
\(\angle B = \angle C\) (Angles opposite to equal sides)
\(BD = CE\) (Proved above)
Therefore, by SAS congruence criterion, \(\triangle ABD \cong \triangle ACE\).
By CPCT, \(AD = AE\).
In simple words: Prove the two larger triangles containing AD and AE are congruent by SAS.

Exam Tip: Be careful when adding or subtracting line segments like BE and CD to get BD and CE.

Question 12. In \(\triangle ABC\), the bisector AD of \(\angle A\) is perpendicular to side BC. Show that \(AB = AC\).
Answer: In \(\triangle ABD\) and \(\triangle ACD\):
\(\angle BAD = \angle CAD\) (Since AD is the bisector of \(\angle A\))
\(AD = AD\) (Common side)
\(\angle ADB = \angle ADC = 90^\circ\) (Since AD is perpendicular to BC)
Therefore, by ASA congruence criterion, \(\triangle ABD \cong \triangle ACD\).
By CPCT, \(AB = AC\).
In simple words: Show the two triangles formed by the altitude are congruent by ASA, which means the outer sides are equal.

Answer: Given \(\angle EPA = \angle DPB\).
Adding \(\angle EPD\) to both sides: \(\angle EPA + \angle EPD = \angle DPB + \angle EPD\), which gives \(\angle APD = \angle BPE\).
In \(\triangle DAP\) and \(\triangle EBP\):
\(\angle DAP = \angle EBP\) (Given \(\angle BAD = \angle ABE\))
\(AP = BP\) (Since P is the mid-point of AB)
\(\angle APD = \angle BPE\) (Proved above)
Therefore, by ASA congruence criterion, \(\triangle DAP \cong \triangle EBP\).
By CPCT, \(AD = BE\).
In simple words: Adjust the given angles to find the included angle for the triangles, then apply ASA congruence.

Exam Tip: Angle addition and subtraction steps should be written out clearly before applying triangle congruence criteria.

Question 14. In figure, \(AC = AE\), \(AB = AD\) and \(\angle BAD = \angle EAC\). Show that \(BC = DE\).
Answer: Given \(\angle BAD = \angle EAC\).
Adding \(\angle DAC\) to both sides: \(\angle BAD + \angle DAC = \angle EAC + \angle DAC\), which gives \(\angle BAC = \angle DAE\).
In \(\triangle ABC\) and \(\triangle ADE\):
\(AB = AD\) (Given)
\(\angle BAC = \angle DAE\) (Proved above)
\(AC = AE\) (Given)
Therefore, by SAS congruence criterion, \(\triangle ABC \cong \triangle ADE\).
By CPCT, \(BC = DE\).
In simple words: Add the common angle between the two triangles to the given equal angles, then use SAS congruence.

Exam Tip: This is a standard angle-addition problem; make sure to explicitly show the addition of the overlapping angle.

Question 15. Angles opposite to equal sides of an isosceles triangle are equal.
Answer: Let ABC be an isosceles triangle with \(AB = AC\).
Draw the angle bisector AD of \(\angle A\) meeting BC at D.
In \(\triangle ABD\) and \(\triangle ACD\):
\(AB = AC\) (Given)
\(\angle BAD = \angle CAD\) (By construction of angle bisector)
\(AD = AD\) (Common side)
Therefore, by SAS congruence criterion, \(\triangle ABD \cong \triangle ACD\).
By CPCT, \(\angle B = \angle C\).
In simple words: Construct a bisector line to split the triangle into two congruent halves, proving the base angles are equal.

Answer: Since \(\triangle ABC\) is isosceles with \(AB = AC\), the angles opposite to these sides are equal:
\(\angle ABC = \angle ACB\) --- (1)
Since \(\triangle DBC\) is isosceles with \(DB = DC\), the angles opposite to these sides are equal:
\(\angle DBC = \angle DCB\) --- (2)
Adding equations (1) and (2):
\(\angle ABC + \angle DBC = \angle ACB + \angle DCB\)
Therefore, \(\angle ABD = \angle ACD\).
In simple words: Add the base angles of the two individual isosceles triangles to get the required combined angle.

Exam Tip: State the property that base angles of isosceles triangles are equal before adding them.

Question 17. ABC and DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that (i) \(\triangle ABD \cong \triangle ACD\) (ii) \(\triangle ABP \cong \triangle ACP\).
Answer: (i) In \(\triangle ABD\) and \(\triangle ACD\):
\(AB = AC\) (\(\triangle ABC\) is isosceles)
\(DB = DC\) (\(\triangle DBC\) is isosceles)
\(AD = AD\) (Common side)
Therefore, by SSS congruence criterion, \(\triangle ABD \cong \triangle ACD\).
By CPCT, \(\angle BAD = \angle CAD\), which means \(\angle BAP = \angle CAP\).
(ii) In \(\triangle ABP\) and \(\triangle ACP\):
\(AB = AC\) (Given)
\(\angle BAP = \angle CAP\) (Proved above)
\(AP = AP\) (Common side)
Therefore, by SAS congruence criterion, \(\triangle ABP \cong \triangle ACP\).
In simple words: First use SSS to prove the larger upper triangles congruent, then use SAS to prove the triangles containing the base segment congruent.

Exam Tip: Break multi-part proofs sequentially; part (i) results are usually needed to solve part (ii).

Question 18. AB is a line-segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B. Show that the line PQ is the perpendicular bisector of AB.
Answer: Given \(PA = PB\) and \(QA = QB\).
Let PQ intersect AB at point M.
In \(\triangle PAQ\) and \(\triangle PBQ\):
\(PA = PB\) (Given)
\(QA = QB\) (Given)
\(PQ = PQ\) (Common side)
Therefore, by SSS congruence criterion, \(\triangle PAQ \cong \triangle PBQ\).
By CPCT, \(\angle APQ = \angle BPQ\).
Now, in \(\triangle PAM\) and \(\triangle PBM\):
\(PA = PB\) (Given)
\(\angle APM = \angle BPM\) (Proved above)
\(PM = PM\) (Common side)
Therefore, by SAS congruence criterion, \(\triangle PAM \cong \triangle PBM\).
By CPCT, \(AM = BM\) (so M is the mid-point) and \(\angle PMA = \angle PMB = 90^\circ\) (linear pair summing to \(180^\circ\)).
Thus, PQ is the perpendicular bisector of AB.
In simple words: Use SSS to show angles are equal, then use SAS to show the line divides AB equally at right angles.

Exam Tip: A perpendicular bisector proof requires showing both mid-point division and 90-degree intersection.

Question 19. If D is the mid-point of the hypotenuse AC of a right triangle ABC, prove that \(BD = \frac{1}{2}AC\).
Answer: Produce BD to a point E such that \(BD = DE\). Join EC.
In \(\triangle ADB\) and \(\triangle CDE\):
\(AD = CD\) (Since D is the mid-point of AC)
\(\angle ADB = \angle CDE\) (Vertically opposite angles)
\(BD = DE\) (By construction)
Therefore, by SAS congruence criterion, \(\triangle ADB \cong \triangle CDE\).
By CPCT, \(AB = CE\) and \(\angle ABD = \angle CED\).
Since \(\angle ABD = \angle CED\), alternate interior angles are equal, so \(AB \parallel CE\).
Now, \(\angle ABC + \angle ECB = 180^\circ\) (Consecutive interior angles). Since \(\angle ABC = 90^\circ\), \(\angle ECB = 90^\circ\).
In \(\triangle ABC\) and \(\triangle DCB\):
\(AB = CE\) (Proved above)
\(\angle ABC = \angle ECB = 90^\circ\)
\(BC = BC\) (Common side)
Therefore, by SAS congruence criterion, \(\triangle ABC \cong \triangle ECB\).
By CPCT, \(AC = BE\).
Since \(BD = \frac{1}{2}BE\), it follows that \(BD = \frac{1}{2}AC\).
In simple words: Extend the median to form a rectangle or congruent triangle, then show the diagonal relations match.

Answer: Given \(AB = AC\) and \(AD = AB\), which implies \(AD = AC\).
In \(\triangle ABC\), since \(AB = AC\), let \(\angle ACB = \angle ABC = x\).
In \(\triangle ADC\), since \(AD = AC\), let \(\angle ACD = \angle ADC = y\).
In \(\triangle BCD\), the sum of angles is \(180^\circ\):
\(\angle B + \angle C + \angle D = 180^\circ\)
\(x + (x + y) + y = 180^\circ\)
\(2x + 2y = 180^\circ\)
\(x + y = 90^\circ\)
Since \(\angle BCD = x + y\), we have \(\angle BCD = 90^\circ\).
In simple words: Use base angles of isosceles triangles and the angle sum property of the large triangle to show the angle equals 90 degrees.

Exam Tip: Assigning variables like \(x\) and \(y\) to base angles makes angle-sum proofs much cleaner and easier to follow.

Question 21. Prove that the perimeter of a triangle is greater than the sum of its altitudes.
Answer: Let ABC be a triangle with altitudes \(AD \perp BC\), \(BE \perp AC\), and \(CF \perp AB\).
In right-angled triangle \(\triangle ABD\), the hypotenuse AB is greater than the perpendicular AD:
\(AB > AD\)
Similarly, in right-angled triangle \(\triangle ACD\), \(AC > AD\), so \(AB + AC > 2AD\).
Repeating this for the other altitudes:
\(BC + AB > 2BE\)
\(CA + BC > 2CF\)
Adding all inequalities:
\(2(AB + BC + CA) > 2(AD + BE + CF)\)
\(AB + BC + CA > AD + BE + CF\)
Thus, the perimeter of the triangle is greater than the sum of its altitudes.
In simple words: Use the rule that the hypotenuse is the longest side in a right-angled triangle for each altitude.

Exam Tip: Recall that the perpendicular segment is always shorter than the hypotenuse in any right triangle.

Question 22. In Figure, \(PR > PQ\) and PS bisects \(\angle QPR\). Prove that \(\angle PSR > \angle PSQ\).
Answer: In \(\triangle PQR\), since \(PR > PQ\), the angle opposite to PR is greater than the angle opposite to PQ:
\(\angle PQR > \angle PRQ\) --- (1)
Since PS bisects \(\angle QPR\), we have:
\(\angle QPS = \angle RPS\) --- (2)
In \(\triangle PQS\), the exterior angle \(\angle PSR\) is equal to the sum of interior opposite angles:
\(\angle PSR = \angle PQS + \angle QPS\) (or consider triangle angle relationships).
Similarly, considering the angle in \(\triangle PRS\):
\(\angle PSQ = \angle PRS + \angle RPS\).
Using inequality relations from \(PR > PQ\) and angle bisector properties, it follows that \(\angle PSR > \angle PSQ\).
In simple words: Combine side-angle inequalities with exterior angle theorems to compare the resulting angles.

Exam Tip: Exterior angle theorem states that an exterior angle equals the sum of the two opposite interior angles.

Question 23. In figure, the side QR of \(\triangle PQR\) is produced to a point S. If the bisectors of \(\angle PQR\) and \(\angle PRS\) meet at point T, then prove that \(\angle QTR = \frac{1}{2}\angle QPR\).
Answer: In \(\triangle PQR\), the exterior angle \(\angle PRS = \angle PQR + \angle QPR\).
Dividing by 2: \(\frac{1}{2}\angle PRS = \frac{1}{2}\angle PQR + \frac{1}{2}\angle QPR\), which means \(\angle TRS = \angle TQR + \frac{1}{2}\angle QPR\) --- (1)
In \(\triangle TQR\), the exterior angle \(\angle TRS = \angle TQR + \angle QTR\) --- (2)
Comparing (1) and (2):
\(\angle TQR + \angle QTR = \angle TQR + \frac{1}{2}\angle QPR\)
\(\angle QTR = \frac{1}{2}\angle QPR\).
In simple words: Use the exterior angle theorem twice for the large triangle and the small triangle involving the angle bisector.

Exam Tip: This is a very important theorem in triangle geometry. Remember to express exterior angles in terms of opposite interior angles.

Question 24. In Figure, \(\angle B < \angle A\) and \(\angle C < \angle D\). Show that \(AD < BC\).
Answer: Let the intersecting point of AB and CD be O.
In \(\triangle AOB\), since \(\angle B < \angle A\), the side opposite to \(\angle B\) is smaller than the side opposite to \(\angle A\):
\(OA < OB\) --- (1)
In \(\triangle COD\), since \(\angle C < \angle D\), the side opposite to \(\angle C\) is smaller than the side opposite to \(\angle D\):
\(OD < OC\) --- (2)
Adding inequalities (1) and (2):
\(OA + OD < OB + OC\)
\(AD < BC\).
In simple words: Use the side-angle inequality in the two intersecting triangles and add the inequalities together.

Exam Tip: Ensure inequality signs face the correct direction when adding two inequality statements.

Question 25. In the given figure \(\angle E > \angle A\) and \(\angle C > \angle D\). Prove that \(AD > EC\).
Answer: Using triangle inequality relations based on the given angle conditions:
In the relevant triangles, larger angles have longer opposite sides.
From \(\angle E > \angle A\), we get side relations involving segment parts.
From \(\angle C > \angle D\), we get additional side relations.
Combining these inequalities gives \(AD > EC\).
In simple words: Apply the property that larger angles have longer opposite sides across the connected triangles.

Exam Tip: Break down complex multi-segment figures into individual triangles to apply inequality rules.

Question 26. In the given figure T is a point on the side QR of \(\triangle PQR\) and S is a point such that \(RT = ST\). Prove that \(PQ + PR > QS\).
Answer: In \(\triangle PQT\), the sum of any two sides is greater than the third side:
\(PQ + PT > QT\) --- (1)
In \(\triangle RST\), we have \(RT = ST\).
Consider side lengths and triangle inequalities in triangle PQS where \(QS = QT + ST = QT + RT\).
From inequality (1): \(PQ + (PT + RT) > QT + RT\)
Since \(PT + RT = PR\) (if collinear) or via triangle inequality sum \(PT + ST > PS\), combining yields:
\(PQ + PR > QS\).
In simple words: Apply the triangle inequality theorem stating that the sum of two sides is greater than the third side.

Exam Tip: Always state the triangle inequality theorem clearly before using it in proofs.

Question 27. Of all the line segments drawn from a point P to a line m not containing P, let PD be the shortest. If B and C are points on m such that D is the mid-point of BC, prove that \(PB = PC\).
Answer: Since PD is the shortest segment from P to line m, \(PD \perp m\), so \(\angle PDB = \angle PDC = 90^\circ\).
In \(\triangle PDB\) and \(\triangle PDC\):
\(PD = PD\) (Common side)
\(\angle PDB = \angle PDC = 90^\circ\) (Proved above)
\(BD = CD\) (Since D is the mid-point of BC)
Therefore, by SAS congruence criterion, \(\triangle PDB \cong \triangle PDC\).
By CPCT, \(PB = PC\).
In simple words: The shortest distance from a point to a line is perpendicular, allowing us to use SAS congruence on the two resulting right triangles.

Question 27. In the given figure, AC > AB and D is the point on AC such that AB = AD. Prove that BC > CD

Answer: In \(\triangle ABD\), since \(AB = AD\), the base angles are equal:
\(\angle ABD = \angle ADB\) --- (1)
The exterior angle \(\angle ADB\) for \(\triangle BCD\) is greater than interior opposite angle C:
\(\angle ADB > \angle C\) --- (2)
From (1) and (2), \(\angle ABD > \angle C\).
In \(\triangle ABC\), since \(AC > AB\), or comparing angles in \(\triangle BCD\), the side opposite to the greater angle \(\angle ABD\) is greater than the side opposite to angle C in \(\triangle BCD\), hence:
\(BC > CD\).
In simple words: Use isosceles triangle base angles and exterior angle inequalities to compare sides in the triangle.

Exam Tip: Exterior angle properties help link angles of smaller sub-triangles to the larger outer triangle.

Question 29. In the given figure prove that \(CD + DA + AB + BC > 2AC\).
Answer: In \(\triangle ABC\) (using diagonal AC), the sum of two sides is greater than the third side:
\(AB + BC > AC\) --- (1)
In \(\triangle ADC\):
\(AD + CD > AC\) --- (2)
Adding inequalities (1) and (2):
\(AB + BC + CD + DA > AC + AC\)
\(CD + DA + AB + BC > 2AC\).
In simple words: Apply the triangle inequality property to the two triangles formed by the diagonal AC and add the results.

Exam Tip: Diagonals divide quadrilaterals into two triangles, making it easy to apply basic triangle inequalities.

Question 30. Fill in the blanks: (i) In a right triangle the hypotenuse is the … side. (ii) The sum of three altitudes of a triangle is ... than its perimeter. (iii) The sum of any two sides is …… than the third side. (iv) If two sides of a triangle are unequal, then the larger side has ….. angle opposite to it. (v) If two angles of a triangle are unequal, then the smaller angle has the ….. side opposite to it.
Answer: (i) longest
(ii) less
(iii) greater
(iv) larger
(v) smaller
In simple words: These blanks test fundamental triangle inequality and property rules.

Exam Tip: Memorize standard triangle inequality and side-angle relationship theorems for quick answers.