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^{°}=30^{}

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 𢈊PD = 75^{°}

Similarly

𢈋PC = 75^{°}

Therefore 𢈍PC = 360^{°}– (75^{°}+75^{°}+60^{°})

or 𢈍PC = 150^{°}

Now in ∆ PDC

PD=PC || As 𢁪PD ≅ 𢁫PC

∴ ∆PDC is an isocles ∆

And ∠PDC = ∠PCD = (180^{°} -150^{°}) /2

Or ∠PDC = ∠PCD = 15^{°}

∴ ∠ DPC=150^{°}

∠ PDC=15 ^{°}

and ∠ PCD=15 ^{°} __Answer __

2nd Case proof for 2nd image and angle calculation are coming soon