Question

what is the value of e square + ed

Answer 1

Answer is

(2)c²

solution

for the ∆ABD...ang(ABD)=ang(BAD)=45°

so the ∆ABD is a isosceles triangle..

so..AD=BD=e

now similarly...BD=CD=d

therefore..e=d=BD

now...

from the ∆ABD...

e²=c²-e²

2e²=c²

..

..

.

from the ∆BDC...

d²+e²=BC²

e²+e²=2e²=BC

now..

from the ∆ABC..

(e+d)²=c²+BC²

=>(e+e)²=c²+2e²

=>4e²-2e²=c²

=>2e²=c²

now ...e²+ed=e²+e²=2e²=c²

Hope this helps you...