Question

In an isosceles triangle ABC with AB=AC,D and E are points on BC such that BE=CD . show that AD=AE.​

Answer 1

Given ABC is an isosceles triangle with AB=AC .D and E are the point on BC such that BE=CD

Given AB=AC

∴∠ABD=∠ACE (opposite angle of sides of a triangle ) ....(1)

Given BE=CD

Then BE−DE=CD−DE

ORBC=CE......................................(2)

In ΔABD and ΔACE

∠ABD=∠ACE ( From 1)

BC=CE (from 2)

AB=AC ( GIven)

∴ΔABD≅ΔACE

So AD=AE [henceproved]

Answer 2

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In an isosceles triangle ABC with AB=AC,D and E are points on BC such that BE=CD. Show that AD=AE.

$\huge{{{{{{{{{{{{{\boxed{\colorbox {aqua} {Answer:-}}}}}}}}}}}}}}}$

~Given ABC is an isosceles triangle with AB=AC

D and E are the point on BC such that BE=CD

$\sf \blue{∴∠ABD=∠ACE (opposite \: angle \: of \: sides \: of \: a \: triangle )}$

Given that, BE=CD

Then BE−DE=CD−DE

ORBC=CE

In ΔABD and ΔACE

∠ABD=∠ACE

BC=CE

AB=AC ( Is given)

$\sf \pink{∴ΔABD≅ΔACE}$

$\sf \red{So \: AD=AE \: [hence \: proved]} \:$

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