### Geometry Sum | ABCD is a square. Prove in each case ∆APD ≅ ∆BPC | Find Angle DPC Angle PCD and Angle PCD |

ABCD is a square and APB is an equilateral triangle.

Prove in each case:

1) 𢁪PD ≅ 𢁫PC

2) Find the angles of 𢁭PC Solution :

First Part

Given �P = Equilateral triangle

R.T.P.

1) 𢁪PD ≅ 𢁫PC

2) Find 𢈍PC, ∠PDC and ∠PCD,

Proof: In 𢁪PD and 𢁫PC

AP = BP | Same side of the equilateral triangle.

AD = BC | Same side of the Square.

and

�P = ∠ DAB - ∠PAB = 90°-60°=300

Similarly

𢈋PC = ∠ ABC - �P = 90°-60°=30°

∴ �P = 𢈋PC

∴ 𢁪PD ≅ 𢁫PC SAS_Proved

2nd Part

In 𢁪PD

AP=AD || As AP = AB (Equilateral Triangle).

We know that �P = 30 °

∴ 𢈊PD = (180°- 30°)/2 || 𢁪PD is an isoscless ∆ and 𢈊PD is one of the base angle.

or 𢈊PD = 150°/2

or …

Prove in each case:

1) 𢁪PD ≅ 𢁫PC

2) Find the angles of 𢁭PC Solution :

First Part

Given �P = Equilateral triangle

R.T.P.

1) 𢁪PD ≅ 𢁫PC

2) Find 𢈍PC, ∠PDC and ∠PCD,

Proof: In 𢁪PD and 𢁫PC

AP = BP | Same side of the equilateral triangle.

AD = BC | Same side of the Square.

and

�P = ∠ DAB - ∠PAB = 90°-60°=300

Similarly

𢈋PC = ∠ ABC - �P = 90°-60°=30°

∴ �P = 𢈋PC

∴ 𢁪PD ≅ 𢁫PC SAS_Proved

2nd Part

In 𢁪PD

AP=AD || As AP = AB (Equilateral Triangle).

We know that �P = 30 °

∴ 𢈊PD = (180°- 30°)/2 || 𢁪PD is an isoscless ∆ and 𢈊PD is one of the base angle.

or 𢈊PD = 150°/2

or …