chivistaelmore

2022-07-23

In $\mathrm{\Delta }ABC,AC>AB.$. The internal angle bisector of $\mathrm{\angle }A$ meets BC at D, and E is the foot of the perpendicular from B onto AD. Suppose $AB=5,BE=4$ and $AE=3$. Find $\left(\frac{AC+AB}{AC-AB}\right)ED$.

nuramaaji2000fh

Expert

Step 1
I use a bit of trigonometry.(I would love to see something without this)
Let $BC=a,CA=b,AB=c$.
By Napiers Analogy:
$\begin{array}{}\text{(1)}& \mathrm{tan}\frac{B-C}{2}=\frac{b-c}{b+c}\mathrm{cot}\left(A/2\right)\end{array}$
Notice that
$\mathrm{sin}\left(A/2\right)=\frac{4}{5}$
$\begin{array}{}\text{(2)}& \mathrm{cot}\left(A/2\right)=?\end{array}$
Step 2
Also $\mathrm{\angle }EDB=90+\frac{C-B}{2}$
$\begin{array}{}\text{(3)}& \mathrm{tan}\mathrm{\angle }EDB=\mathrm{tan}\left(90+\frac{C-B}{2}\right)=\mathrm{cot}\frac{B-C}{2}=\frac{4}{ED}\end{array}$

Jaxon Hamilton

Expert

Step 1
This proof can probably be simplified or made more natural, but here is my take:

Construct CG parallel to FB as shown above. We have:

The last ratio is by Angle Bisector Theorem.
Step 2
Hence: $\frac{AC+AB}{AC-AB}=\frac{AC}{CF}+\frac{AB}{CF}=\frac{AB}{CF}\left(1+\frac{AC}{AB}\right)=\frac{AF}{CF}\left(1+\frac{DG}{ED}\right)=\frac{AE}{EG}\frac{EG}{ED}=\frac{AE}{ED}$
Giving the result: $\left(\frac{AC+AB}{AC-AB}\right)ED=AE$

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