Question

In the adjacent figure ABCD is square and ∆ APB is an equilateral triangle . Prove that ∆ APD ≈ ∆ BPC .​

Answer 1

Answer:

there are three conditions to congurenent a triangle.they are..

1. AD=BC

2. angle DAP= angleCBP

3.AP=BP

by SAS (side angle side) congurence condition ......

∆APD congurent to ∆BPC

Answer 2

Step-by-step explanation:

$\huge\mathbb\red{Given~:-}$

ABCD is a square

angle DAB = angle CBA = 90°

∆ APB is an equivalateral triangle

angle PAB = angle PBA = 60°

$\huge\mathbb\red{Required~to~prove~:-}$

∆ APB = ∆ BPC

$\huge\mathbb\red{Proof~:-}$

from the diagram

angle DAP = angle DAB - angle PAB

angle DAP = 90° - 60°

angle DAP = 30°

_______________________________

angle CBP = angle CBA - angle PBA

angle CBP = 90° - 60°

angle CBP = 30°

:. angle DAP = angle CBP → 1

In ∆ APD and ∆ BPC

AD = BC $\huge\green{(sides~of~square)}$

angle DAP = angle CBP $\huge\pink{(from~eqn~1)}$

AP = BP $\huge\blue{(sides~of~equilateral~traingle)}$

By SAS congruent rule

∆ APD ≈ ∆ BPC

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