Question

in the adjacent figure ABCD is a square and∆APB is an equilateral triangle.Prove that ∆APD is congruent to ∆BPC
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Answer 1

Answer:

Step-by-step explanation:

simple in ADP and BPC

AD =BC

AP=BP

angle DAP=ANGLE CBP = 30

SAS RULE FOR CONGRUENCE

Answer 2

Answer:∆APD is congruent to ∆BPC.

Step-by-step explanation:

• Given: ABCD is a square. ∆APB is an equilateral triangle and P lies outside the square.

• To prove: ∆APD is congruent to ∆BPC.

• Proof: In ∆APD and ∆BPC,

AD=BC (Sides of the square)

AP=BP (Sides of the equilateral triangle)

• angleDAP=(angleDAB+angleBAP)

= 90° + 60° = 150°

angleCBP=(angleCBA+angleABP)

= 90° + 60° = 150°

• angleDAP=angleCBP

• Therefore, by SAS

∆APD is congruent to ∆BPC.