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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: 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.

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  • $\begingroup$ For a hint, consider the fact that $AD=\frac{AB+AC}{2}$. $\endgroup$ – nickgard Jul 2 '17 at 11:19
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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, $|CE|:|EB|=|CD|:|DB'|=1:1$.

$\qquad$enter image description here

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  • $\begingroup$ Fantastic! thanks :) $\endgroup$ – y12 Jul 4 '17 at 16:50
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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!

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  • $\begingroup$ 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 $\endgroup$ – Ido Sarig Jul 14 at 14:50
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    $\begingroup$ @Ido Sarig I fixed. Thank you! The essence is the same. $\endgroup$ – Michael Rozenberg Jul 14 at 15:23

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