Question

ABC is an isosceles triangle such that AB=AC. D is The mid point of AC. A circle is drawn taking BD as diameter wch intersects AB at point E. Prove that AC=4AE.

Answer 1

Given : ABC is an isosceles triangle such that AB=AC. D is The mid point of AC.

A circle is drawn taking BD as diameter which intersects AB at point E.  

To Find : Prove that AC=4AE.

Solution:

AB = AC

D is The mid point of AC

=> AD = AC/2 = AB/2

BD is diameter

=> ∠BED = 90°

=> ∠AED = 90°

DE ⊥ AB

=> DE² = AE * BE

 => DE² = AE * (AB - AE)

 => DE² = AE *  AB - AE²

DE² = AD² - AE²

=> DE² = (AB/2)² - AE²

Equating DE²

(AB/2)² - AE² = AE *  AB - AE²

=> (AB/2)²  = AE *  AB

=> AB²/4 =  AE *  AB

=> AB = 4AE

=> AC = 4AE

QED

Hence proved

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