In the figure, $AB \bot CK$, $\angle B = 2\angle A$, $I$ and $J$ are the mid-points of $AB$ and $BC$. Prove that $IK = \frac{1}{2} BC$.
It can easily be solved with trigonometry by letting $\angle CAB = \theta, BC = \sin \theta$ and show that both $IK$ and $\frac{1}{2}BC$ are $\frac{\sin \theta}{2}$, however, I am interested in a pure geometry solution if there exists any, thanks in advance!
Reflect point $B$ with respect to segment $CK$ to obtain point $B’$ as shown in figure below
Since $\angle KB’C =2\angle KAC$, triangle $CB’A$ is isosceles with $AB’=B’C$.
$BC=B’C=AB’=AB-2BK$
$IK=IB-BK=\frac{1}{2}AB-BK$
The right angle at $K$ implies (by Thales' Theorem) that $K$ lies on a semicircle with center $J$ and diameter $\overline{BC}$; hence, $|JK|=|JB|=|JC|=\frac12|BC|$. Also, midsegment $\overline{IJ}$ of $\triangle ABC$ is parallel to $\overline{AC}$.
A little angle chasing yields $\angle KIJ=\angle IJK$, so that $|IK|=|JK|$. $\square$
Join point J and K. Since j is the midpoint of hypotenuse
=> Segment JK = BJ = CJ = BC/2 (given)
=> Angle (BCK) = Angle (90° - 2A)
=> Angle (JKC) = Angle (90° - 2A) ( since ∆JKC is issoceles)
=> Angle (IJK) = 180° - (90° + 90 - 2A) - A ( angle sum property of ∆)
=> Angle (IJK) = Angle(A)
=> IK = JK (∆ IJK is issoceles)
=> IK = JK = BC/2 Proved
