Question

In the following diagram, ABCD is a square and APB is an equilateral triangle.

(i) Prove that: ΔAPD≅ ΔBPC
(ii) Find the angles of ΔDPC.

Answer 1

Given: ABCD is a Square and ΔAPB is an equilateral triangle.
We need to

(i) Prove that: ΔAPD≅ ΔBPC

(ii) Find the angles of ΔDPC

Proof:

Since AB side is present in both square & equilateral triangle

AP = PB = AB =AD = CD = BC

(i) In ΔBPC,

BP  = BC

∴ ∠BPC = ∠PCB

∠BPC + ∠PCB + 30° = 180°

∠BPC  + ∠BPC  = 150°

2∠BPC = ` (150°)/2 = 75°`

∴ ∠BPC = ∠PCB = 75°

∠ADP = ∠DPA = 75°    ...[C.P.C.T.C]

(ii) In DPC

∠DCP =  90° - 75° =  15° 

∠PDC = 90° - 75° =  15° 

∠DPC = 180° - (15° + 15°)

∠DPC = 150°