Question

In ∆ABC, the bisector of exterior angle of angle B and and angle C meets at P. Prove that Angle BPC=90°-A/2​

Answer 1

Answer:

Step-by-step explanation:

a+b+c=180

a+b=180-90/2

a+b=90-a/2

Answer 2

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 :-

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