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) ∆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^°=30⁰
Similarly
∠BPC = ∠ ABC - ∠ABP = 90^°-60^°=30^°
∴ ∠DAP = ∠BPC
∴ ∆APD ≅ ∆BPC SAS_Proved

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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