MHB How Do You Prove Equidistance in This Isosceles Right Triangle Geometry Problem?

[anemone]

Here is this week's POTW:

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On the sides $AC$ and $BC$ of an isosceles right-angled triangle $ABC$, points $D$ and $E$ are chosen such that $|CD|=|CE|$. The perpendiculars from $C$ and $D$ on $AE$ intersect the hypotenuse $AB$ at $L$ and $K$ respectively. Prove that $|LK|=|LB|$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to Opalg for his correct solution, which you can find below::)

Spoiler

[TIKZ][scale=0.5]
\coordinate [label=left: $A$] (A) at (0,10) ;
\coordinate [label=below: $B$] (B) at (10,0) ;
\coordinate [label=below: $C$] (C) at (0,0) ;
\coordinate [label=above left: $D$] (D) at (0,4) ;
\coordinate [label=below: $E$] (E) at (4,0) ;
\coordinate [label=above right: $K$] (K) at (4.29,5.71) ;
\coordinate [label=above right: $L$] (L) at (7.14,2.86) ;
\coordinate [label=below: $P$] (P) at (-10,0) ;
\draw [very thick](C) -- (A) -- (B) -- cycle ;
\draw (B) -- (P) -- (K) ;
\draw (A) -- (E) ;
\draw (C) -- (L) ;
[/TIKZ]

Extend the lines $KD$ and $BC$ so that they meet at $P$. Since $KP$ is perpendicular to $AE$, and $PC$ is perpendicular to $AC$, it follows that the angles $KPC$ and $EAC$ are equal. Thus the right-angled triangles $EAC$ and $DPC$ have the same angles. But $|CE| = |CD|$, so those two triangles are in fact congruent. Therefore $|PC| = |CB|$, so that $C$ is the midpoint of $PB$.

The lines $CL$ and $PK$ are parallel, which means that the triangles $CLB$ and $PKB$ are similar. Since $C$ is the midpoint of $PB$, it follows that $L$ is the midpoint of $KB$. Thus $|LK| = |LB|$.