# Finding the length of angle bisectors

• Aug 28th 2009, 05:42 PM
Mp5xm8
Finding the length of angle bisectors
I can't seem to figure out the answer to these two problems, I have been trying to work on them with no luck. Could someone please explain to me how to do them, and which theorems are relevant to them. Thanks. (Cool)

4. In Triangle ABC , suppose that the bisector of angle B meets side AC at point E. If AB = 12, BC = 14, and AC = 18,
find AE and EC.

5. Given right triangle ABC with right angle at C, altitude CD is drawn to the hypotenuse of the triangle. If AD =
12 ,and DB = 4 3, find AC, CB, and CD.
• Aug 28th 2009, 06:48 PM
ynj
Quote:

Originally Posted by Mp5xm8
I can't seem to figure out the answer to these two problems, I have been trying to work on them with no luck. Could someone please explain to me how to do them, and which theorems are relevant to them. Thanks. (Cool)

4. In Triangle ABC , suppose that the bisector of angle B meets side AC at point E. If AB = 12, BC = 14, and AC = 18,
find AE and EC.

5. Given right triangle ABC with right angle at C, altitude CD is drawn to the hypotenuse of the triangle. If AD =
12 ,and DB = 4 3, find AC, CB, and CD.

for 4 you must use a important theorem of angle bisector that
$\displaystyle \frac{AB}{BC}=\frac{AE}{EC}$always holds.
This proof is a little bit complex..
Construct a line $\displaystyle CF\parallel BE$such that the extension line of $\displaystyle AB$intersect with $\displaystyle CF$at $\displaystyle F$.
So $\displaystyle \angle ABE=\angle AFC=\angle EBC=\angle BCF$, so $\displaystyle BC=BF$. But since$\displaystyle CF\parallel BE$, $\displaystyle \frac{AB}{BF}=\frac{AE}{EC}$, so $\displaystyle \frac{AB}{BC}=\frac{AE}{EC}$ since $\displaystyle BC=BF$
for 5 ,you may use the important theorem called projective theorem which says that $\displaystyle CD^2=AD\cdot DB,AC^2=AD\cdot AB, BC^2=BD\cdot AB$, this can be directly deduced from the fact that $\displaystyle \triangle ABC\sim\triangle ACD\sim\triangle CBD$