**Problem-9: In the triangle ABC, AB>AC and the bisectors of the ∠B and ∠C intersect at the point P. Prove that PB>PC.**

**Particular enunciation**: Given that, in the triangle ABC, AB>AC and the bisectors of the ∠B and ∠C intersect at the point P. We have to prove that PB>PC.

**Proof**: Since, BP is the bisector of ∠B.

∴∠PBC=½∠ABC

CP is the bisector of ∠C.

∴∠PCB = ½ ∠ACB

We know that, in a triangle the angle opposite the greater side is greater than the angle opposite the smaller side.

Since AB>AC

∴ ∠ACB>∠ABC

⟹½∠ACB> ½∠ABC

⟹∠PCB>∠PBC

⟹PB>PC

∴ PB>PC.(Proved)

**Problem -10.ABC is an isosceles triangle and AB=AC. The side BC is extend up to D. Prove that AD>AB.**

**General enunciation**: ABC is an isosceles triangle and AB=AC. The side BC is extend up to D. Prove that AD>AB.

**Particular enunciation**: Given that, the ∆ABC is an isosceles triangle and AB=AC. The side BC is extended up to D. We have to prove that AD>AB.

**Proof**: Given that in ∆ ABC,

AB=AC

∴ ∠ACB=∠ABC [∵ opposite angles of equal arms are also equal]

Again, ∠ACB is the external angle of ∆ ACD

∴ ∠ACB>∠ADC

⟹ ∠ABC>∠ADC

⟹ ∠ABD>∠ADB

⟹AD>AB

∴ AD>AB (proved)

**Problem – 11: In the quadrilateral ABCD, AB = AD, BC = CD and CD> AD. Prove that ∠DAB > ∠BCD.**

**General enunciation**: In the quadrilateral ABCD, AB = AD, BC = CD and CD> AD. Prove that ∠DAB > ∠BCD.

**Particular enunciation**: In the quadrilateral ABCD, AB = AD, BC = CD and CD> AD. We have to prove that ∠DAB > ∠BCD.

Construction: Join A and C.

**Proof**: We know that, in a triangle the angle opposite the greater side is greater than the angle opposite the smaller side.

In the triangle ∆CAD, CD > AD

∴ ∠DAC > ∠ACD ———————————— (1)

Again, in the triangle ∆ ABC, BC > AB

∴ ∠BAC > ∠ACB ————————————- (2)

Adding equation (1) and (2) we get,

∠DAC + ∠BAC > ∠ACD + ∠ACB

⟹ ∠DAB > ∠BCD (Proved)

**Problem-12: In the triangle ∆ABC, AB = AC and D is any point on BC. Prove that AB > AD.**

**General enunciation**: In the triangle ∆ABC, AB = AC and D is any point on BC. Prove that AB > AD.

**Particular enunciation**: Given that, in the triangle ∆ABC, AB = AC and D is any point on BC. We have to prove that AB > AD.

Proof: In the ∆ ABC, AB = AC

∴ ∠ACB = ∠ABC [∵ opposite angles of equal arms are also equal]

⟹ ∠ACD= ∠ABD

Now ∠ADB is the exterior angle of ∆ADC

∴ ∠ADB > ∠ACD

⟹ ∠ADB > ∠ABD

⟹ AB > AD (Proved)

**Problem – 13: In the triangle ∆ABC, AB ⊥ AC and D is any point on AC. Prove that BC > BD.**

**General enunciation**: In the triangle ∆ABC, AB ⊥ AC and D is any point on AC. Prove that BC > BD.

**Particular enunciation**: In the triangle ∆ABC, AB ⊥ AC and D is any point on AC. We have to prove that BC > BD.

Construction: Join B and D.

Proof: In ∆ABC, AB ⊥ AC

∴ ∠BAC = 90^{0}, BC be the hypotenuse.

∴ ∠BCA is an acute angle

∴ ∠BAC > ∠BCA

Again, ∠BDC is an exterior angle of ∆ABD.

∴ ∠BDC > ∠BAD

⟹ ∠BDC > ∠BAC

⟹ ∠BDC > ∠BCA

⟹ ∠BDC > ∠BCD

⟹ BC > BD (Proved)

**Problem-14: Prove that the hypotenuse of a right angled triangle is the greatest side.**

** General enunciation**: We have to prove that the hypotenuse of a right angled triangle is the greatest side.

**Particular enunciation**: Let ∆ABC be right angled triangle in which ∠ABC = right angle or 90^{0} and AC be the hypotenuse. It is required to prove that AC be the greatest side.

**Proof**: We know that, in a right angled triangle one angle is right angle and the other two angles are acute angles.

i.e., ∠ABC = 1 right angle = 90^{0}

∴ ∠ABC> ∠ACB

∴ AC>AB [∵the side opposite to the greater angle is greater.]

Again, ∠ABC> ∠BAC

⟹ AC>BC

∴ Side AC be greater than the sides AB and BC.

Therefore, AC be the greatest side. (Proved)

**Problem-15: Prove that the angle opposite the greatest side of a triangle is also the greatest angle of that triangle.**

** General enunciation**: We have to prove that the angle opposite the greatest side of a triangle is also the greatest angle of that triangle.

**Particular enunciation**: Let the greatest side of ∆ABC is AC. We have to prove that ∠ABC is the greatest angle of the triangle.

**Proof**: In triangle ABC, AC>AB

∴ ∠ABC> ∠ACB [∵the angle opposite to the greater side is greater.]

Again, AC>BC

⟹∠ABC> ∠BAC

Hence ∠ABC be greater than ∠ACB and ∠BAC.

Therefore, ∠ABC be the greatest angle of the triangle.