# Geometry problem - perpendicular line to angle bisector

Here is my question:

Let $\triangle ABC$ and let $K,L$ be midpoints of $AC$ and $AB$ respectively. Let $D$ be some point on $AC$ (between $K$ and $C$) such that $KD=AL$.

Show that the perpendicular line from $D$ to the angle bisector of $A$ halves $BC$.

First of all, I've drawn the following drawing:

I tried to connect $KE$ and $KL$ and to prove that $KE$ is parallel to $AB$, but to no avail.

Please give a hint, I find this question very hard.

• For a hint, consider the fact that $AD=\frac{AB+AC}{2}$. – nickgard Jul 2 '17 at 11:19

Let $B'\in AC$ such that $|AB'|=|AB|$ and $B',C$ are on the same side of $A$. Then $D$ is the midpoint of $B'C$. Furthermore, $B'$ is $B$ mirrored across the internal angle bisector $AF$. Therefore $DF\|B'B$. Due to the intercept theorem, $|CE|:|EB|=|CD|:|DB'|=1:1$.

$\qquad$

• Fantastic! thanks :) – y12 Jul 4 '17 at 16:50

Let $$DF\cap AB=\{M\}$$ and $$F'\in EM$$ such that $$BF'||AC$$.

Thus, $$AD=AM$$ and since $$DK=AL=LB$$ and $$AK=KC$$, we obtain $$DC=MB$$.

But $$\measuredangle ADM=\measuredangle BMF'=\measuredangle BF'M$$, which gives $$BF'=BM$$.

Also $$\measuredangle DEC=\measuredangle BEF'.$$

Thus, $$\Delta CDE\cong \Delta BF'E$$, which gives $$CE=BE$$ and we are done!

• There's some confusion here between the F you define and the F given in the drawing of the problem. Also it should be BF||AC, not BM – Ido Sarig Jul 14 at 14:50
• @Ido Sarig I fixed. Thank you! The essence is the same. – Michael Rozenberg Jul 14 at 15:23