Problem 5

Let

ABC

be a triangle with

\angle BCA=90^\circ

, and let D be the foot of the altitude from C. Let X be interior to CD. Let K lie on AX with

BK=BC

, and L lie on BX with

AL=AC

. Let

M=AL\cap BK

. Prove

MK=ML

Step 3 of 5: Put L and K on right-angle circles

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In plain words

The angle equality makes R,A,D,L cyclic, and the resulting right angle turns RL and RK into computable lengths.

R,A,D,L\text{ cyclic},\qquad RL^2=AR^2-AC^2

Detailed analysis

The equality $\angle ALD=\angle ARD$ puts $R,A,D,L$ on a circle. Therefore $\angle RLA=90^\circ$ and $RL^2=AR^2-AL^2=AR^2-AC^2$ . By the symmetric argument for K, $\angle RKB=90^\circ$ and $RK^2=BR^2-BK^2=BR^2-BC^2$ .