Question

c) In the figure alongside, BP and CP are the angular bisectors
of the exterior angles BCD and CBE of triangle ABC. Prove that
angle BPC = 90°- angleA/2​

Answer 1

Solution :-

→ Ext.(∠BCD) = ∠A + ∠B { Exterior angle is equal to sum of opposite interior angles . }

So,

→ ∠BCP = (1/2)Ext.(∠BCD) { CP is angular bisector of Ext.(∠BCD) .}

→ ∠BCP = (1/2)[∠A + ∠B] -------- Eqn.(1)

similarly,

→ Ext.(∠CBE) = ∠A + ∠C { same above }

→ ∠CBP = (1/2)Ext.(∠CBE) { BP is angular bisector . }

→ ∠CBP = (1/2)[∠A + ∠C] --------- Eqn.(2)

Now, in ∆BPC we have,

→ ∠BCP + ∠CBP + ∠BPC = 180° { By angle sum property. }

putting values from Eqn.(1) and Eqn.(2),

→ (1/2)(∠A + ∠B) + (1/2)(∠A + ∠C) + ∠BPC = 180°

→ (1/2)(∠A + ∠A + ∠B + ∠C) + ∠BPC = 180°

putting ∠A + ∠B + ∠C = 180° ,

→ (1/2)(∠A + 180°) + ∠BPC = 180°

→ (1/2)∠A + 90° + ∠BPC = 180°

→ ∠BPC = 180° - 90° - (1/2)∠A

→ ∠BPC = [90° - (1/2)∠A] (proved.)

Learn more :-

In the figure ∠ MNP = 90°, ∠ MQN = 90°, , MQ = 12 , QP = 3 then find NQ .

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show that AB2 = AD.AC

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Answer 2

Answer:

Answer:

Step-by-step explanation:

From the figure attached,

CP is an angle bisector of angle BCD and BP is the angle bisector of angle CBE

Therefore, m∠DCP ≅ m∠BCP

and m∠PBE ≅ m∠PBC

m∠A + m∠CBA = m∠BCD

m∠A + (180° - m∠CBE) = m∠BCD

m∠A + 180° = m∠CBE + m∠BCD

m∠A + 180° = 2(m∠PCB) + 2(m∠PBC) [Since m∠CBE = 2m∠PCB and m∠BCD = 2(m∠PBC)

m∠A + 180° = 2(m∠PCB + m∠PBC)

m∠A + 180° = 2(180° - m∠BPC) [Since m∠PCB + m∠PBC + m∠BPC = 180°]

[\frac{1}{2}(m\angle A)]+90=180 - m\angle BPC[

2

1

(m∠A)]+90=180−m∠BPC

m∠BPC = 180 - [\frac{1}{2}(m\angle A)]-90[

2

1

(m∠A)]−90

m∠BPC = 90 - \frac{1}{2}m\angle A

2

1

m∠A