Question

in the fig 7.36,AB=AC .D is a point on AC and E is a point on AB such that AD = DE=EC = BC prove that∆A:∆B=1:3.

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Answer 1

GIVEN:

AB = AC

AD = DE = EC = BC

 

TO PROVE:

<A : <B = 1:3

PROOF:

Since AD = DE

=> < A = < AED = x ( isosceles triangle)…….(1)

=> < CDE = x + x = 2x ( exterior angle = the sum of 2 opposite interior angles)

But, DE =CE ( given)

So, < CDE = < DCE = 2x

=> < BEC = 2x + x = 3x ( exterior angle of triangle ACE )

And CE = CB ( given)

=> < CEB = < CBE = 3x …………(2)

By (1) & (2)

< A / <B = x/3x = 1/3

<A : <B = 1 : 3

PROVEN:  

<A = <B = 1:3