Question

. In an acute angled triangle ABC, the internal bisector of angle A meets base BC at point D. DE I AB and DF I AC; show that AEF is an isosceles triangle.​

Answer 1

$\sf\small{ \underline{\underline\red{ \pmb{Given:-}}}}$

ABC is an acute angled triangle, the internal bisector of angle A meets base BC at point D.

• DE ⊥ AB
• DF ⊥ AC

$\sf\small{ \underline{\underline\red{ \pmb{To \: Find:-}}}}$

• AEF is an isosceles triangle.

$\sf\small{ \underline{\underline\red{ \pmb{ Solution:-}}}}$

$\rm \: In \: △ADE \: and \: △ADF,$

$\rm \: ∠DAE=∠DAF \: \: \: (AD \: bisects \: ∠A)$

$\rm \: AD=AD \: \: \: \: \: (Common)$

$\rm \: ∠AED=∠AFD \: \: (Each \: 90° )$

$\sf$

$\rm \: △ADE≅△ADF \: \: \: \: (ASA \: \: postulate)$

$\rm \: AE = AF \: \: \: (By \: \: cpct)$

$\boxed{ \red{\rm \therefore \: △AFE \: is \: an \: isosceles \: triangle.}}$