# Geometry problem without trigonometry 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!

• If you see a right angle, there must be a circle somewhere. Aug 4 at 1:16
• Why is the point $J$ part of the problem? Aug 4 at 1:28
• It shouldn't be. I just realized that it is useless. Or is the problem setter trying to give a hint? I am not sure. Aug 4 at 1:32

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