MHB Proving BC=AE+BE in $\triangle ABC$

[Albert1]

$\triangle ABC,\, AB=AC ,\,\, \angle A=100^o$
the angle bisectoer of $\angle B $ intersects AC at point E
prove BC=AE+BE

 

[kaliprasad]

I could not draw the figure

but

Spoiler

in $\triangle$ ABE we have $\angle ABE = 20^o$

so $\frac{AE}{BE}= \frac{\sin \; 20^0}{\sin \; 100^0}$

similarly in

in $\triangle$ BCE we have $\angle BEC = 120^o$

so $\frac{BC}{BE}= \frac{\sin \; 120^0}{\sin \; 40^0}$

as we need to prove

BC = AE + BE

or $ \frac{BC}{BE} = 1 + \frac{AE}{BE}$

or $\frac{\sin \; 120^0}{\sin \; 40 ^0}= 1 + \frac{\sin \; 20^0}{\sin \; 100^0} $

we have RHS
= $1 + \frac{\sin \; 20^0}{\sin \; 100^0} $
= $ \frac{\sin \; 20^0+ \sin \; (100)^0} {\sin \; 100^0} $
= $ \frac{\sin \; 20^0+ \sin \; (100)^0} {\sin \; (180-100)^0} $
= $ \frac{\sin \; 20^0+ \sin \; (100)^0} {\sin \;80^0} $
= $ \frac{2 \sin \; 60^0 \cos \;40 ^0} {2 \cos \;40^0\sin \;40^0} $
=$ \frac{ \sin \; 60^0 } {sin \;40^0} $
= $ \frac{ \sin \;(180- 60)^0 } {sin \;40^0} $
= $ \frac{ \sin \;120^0 } {sin \;40^0} $
=LHS

 

[Albert1]

Spoiler

from the diagram it is easy to see that :BC=AE+BE
View attachment 2536