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
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
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