Apr 29, 2015

# 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
GivenABP = Equilateral triangle
R.T.P.
1)APD ≅BPC
2) Find DPC,PDC andPCD,
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 || AsAPD ≅BPC
PDC is an isocles
And PDC = PCD = (180° -150°) /2
Or PDC = PCD = 15°
DPC=150°
PDC=15 °

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

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