ABCD is a square and APB is an equilateral triangle.
Prove in each case:
  1) &#x2206APD ≅ &#x2206BPC
   2) Find the angles of &#x2206DPC

Solution :
First Part
Given &#x2206ABP = Equilateral triangle
R.T.P.
1) &#x2206APD ≅ &#x2206BPC
   2) Find &#x2220DPC, &#x2220PDC and &#x2220PCD,
Proof: In &#x2206APD and &#x2206BPC

AP = BP | Same side of the equilateral triangle.
AD = BC | Same side of the Square.
and
&#x2220DAP = &#x2220 DAB – &#x2220PAB = 90°-60°=300
Similarly
&#x2220BPC = &#x2220 ABC – &#x2220ABP = 90°-60°=30°
&#x2220DAP = &#x2220BPC
&#x2206APD ≅ &#x2206BPC SAS_Proved


2nd Part

In &#x2206APD
AP=AD || As AP = AB (Equilateral Triangle).
We know that &#x2220DAP = 30 °
&#x2220APD = (180°– 30°)/2 || &#x2206APD is an isoscless &#x2206 and &#x2220APD is one of the base angle.
or &#x2220APD = 150°/2
or &#x2220APD = 75°
Similarly
&#x2220BPC = 75°
Therefore &#x2220DPC = 360°– (75°+75°+60°)
or &#x2220DPC = 150°
Now in &#x2206 PDC
PD=PC || As &#x2206APD ≅ &#x2206BPC
&#x2206PDC is an isocles &#x2206
And &#x2220PDC = &#x2220PCD = (180° -150°) /2
Or &#x2220PDC = &#x2220PCD = 15°
&#x2220 DPC=150°
&#x2220 PDC=15 °
and &#x2220 PCD=15 ° Answer

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

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