Math, asked by sambitghatak2008, 19 hours ago

. 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.​

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Answered by ShiningBlossom
4

\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.}}

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