ABCD is a square and APB is an equilateral triangle.

Prove in each case:

1) ∆APD ≅ ∆BPC

2) Find the angles of ∆DPC

Solution :

First Part

Given ∆ABP = Equilateral triangle

R.T.P.

1) ∆APD ≅ ∆BPC

2) Find ∠DPC, ∠PDC and ∠PCD,

Proof:
In ∆APD and ∆BPC

AP = BP | Same side of the equilateral triangle.

AD = BC | Same side of the Square.

and

∠DAP = ∠ DAB - ∠PAB = 90^{°}-60^{°}=30^{0}

Similarly

∠BPC = ∠ ABC - ∠ABP = 90^{°}-60^{°}=30^{°}

∴ ∠DAP = ∠BPC

∴ ∆APD ≅ ∆BPC __ SAS_Proved __

2nd Part

In ∆APD

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

We know that ∠DAP = 30 ^{°}

∴ ∠APD = (180^{°}- 30^{°})/2 || ∆APD is an isoscless ∆ and ∠APD is one of the base angle.

or ∠APD = 150^{°}/2

or ∠APD = 75^{°}

Similarly

∠BPC = 75^{°}

Therefore ∠DPC = 360^{°}- (75^{°}+75^{°}+60^{°})

or ∠DPC = 150^{°}

Now in ∆ PDC

PD=PC || As ∆APD ≅ ∆BPC

∴ ∆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