Question

Triangle ACB is iscoceles triangle with a Angle AB=AC and DE is parallel to BC. Show that Angle ADE= Angle AED.

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Answer 1

Step-by-step explanation:

REF.image

To prove : AB×EF=AD×EC

⇒

EC

AB

=

EF

AD

Proof :

AB=AC (∵ ABC is isosceles)

∴∠B=∠C (angles opposite to equal sides are equal) - (1)

In ΔABD and ΔECF

∠ABD=∠ECF (from (1))

∠ADB=∠EFC (Both are 90

∘

)

Using AA similarity

ΔADB∼ΔECF

⇒

EC

AB

=

EF

AD

⇒AB×EF=AD×EC

∴ Hence proved.

solution

Answer 2

Step-by-step explanation:

REF.image

To prove : AB×EF=AD×EC

⇒

EC

AB

=

EF

AD

Proof :

AB=AC (∵ ABC is isosceles)

∴∠B=∠C (angles opposite to equal sides are equal) - (1)

In ΔABD and ΔECF

∠ABD=∠ECF (from (1))

∠ADB=∠EFC (Both are 90

∘

)

Using AA similarity

ΔADB∼ΔECF

⇒

EC

AB

=

EF

AD

⇒AB×EF=AD×EC