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Question

ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that ΔABEΔACF
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Solution
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In ABE and ACF, we have

AEB=AFC [Since Each =90o]

BAE=CAF [Common]

and AB=AC [Given]

By AAS criterion of congruence, we have

ABEACF

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463834.jpg

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