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ABC is an isosceles triangle with …

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ABC is an isosceles triangle with AB=AC. D is the midpoint of AC. Taking BD as diameter a circle is drawn which meets AB at E. Prove that AE=1/4AC

  • 1 answers

Naveen Sharma 7 years ago

Ans.

Given: ABC is isosceles triangle. AB = AC. D is mid-point of AC.

To prove : {tex}AE = {1\over 4}AC{/tex}

Proof:

AB = AC …………(1)   (Given)

AD = AC/2   ……………(3) (d is midpoint)

As AD is tangent to the triangle and AEB is secant then,

{tex}=> AE \times AB = AD^2{/tex}

{tex}=> AE \times AC = ({AC\over 2})^2{/tex}

{tex}=> AE \times AC = {AC^2\over 4}{/tex}

{tex}=> AE = {AC\over 4}{/tex}

Hence Proved.

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