In △ A B C, ∠ B < ∠ C, and B D and C...

Rebecca Villa

Rebecca Villa

Answered

2022-07-09

In A B C, B < C, and B D and C E are angle bisectors. D is on A C and E is on A B. Prove that C E < B D.

Using the fact that B < C, I got that A B > A C and B F > C F (where F is the intersection of the two angle bisectors).
I then expressed C E = C F + F E and B D = B F + D F. Since C F < B F, we only have to prove that F E < D F. I tried to use the angle bisector theorem to get some relations that might help me prove that, but here's where I got stuck. Could someone please help me out?

Answer & Explanation

Jenna Farmer

Jenna Farmer

Expert

2022-07-10Added 17 answers

Use theorem of sines on triangles A B D and A C E to obtain:
B D sin A = A B sin ( A + B 2 ) , C E sin A = A C sin ( A + C 2 ) .
Combining these two yields:
B D C E = A B A C sin ( A + C 2 ) sin ( A + B 2 )
and the rest should be easy.

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