Let ABCD be a rectangle such that AB = $ \sqrt {2} BC $. Let E be a point on the semicircle with diameter AB, as shown in the following figure. Let K and L be the intersections of AB with ED and EC, respectively. If AK = 2cm and BL = 9 cm, calculate, in cm, the length of segment KL.
I tried to do this question in so many ways (Pythagoras, Similar Triangles...), but I always I found 0=0 or 1=1. Can someone help me? The answer is KL=6 cm.
Let $H$ be the projection of $E$ onto $AB$. Set $x=AB$ and $y=EH$. We have then $BC=AD=x/\sqrt2$ and (from the similarity of $EHK$ and $AKD$) $KH=AK\cdot EH/AD=2\sqrt2y/x$, so that: $$AH=2\Big(1+\sqrt2{y\over x}\Big).$$ Analogously: $$BH=9\Big(1+\sqrt2{y\over x}\Big)$$ and $$ \tag{1} x=AH+BH=11\Big(1+\sqrt2{y\over x}\Big). $$ From the Intersecting chords theorem we also have: $$ \tag{2} y=\sqrt{AH\cdot BH}=3\sqrt2\Big(1+\sqrt2{y\over x}\Big). $$ Dividing $(2)$ by $(1)$ we can find $y/x$ and substituting that into $(1)$ gives $x=17$.