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 1 of 5: Use the altitude-foot similarity

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In plain words

The right triangle and the altitude foot convert the length condition AL=AC into a useful product on AB.

AL^2=AC^2=AD\cdot AB

Detailed analysis

Since $AL=AC$ and the altitude theorem gives $AC^2=AD\cdot AB$ , we have $AL^2=AD\cdot AB$ . Thus triangles $ALD$ and $ABL$ are similar, so $\angle ALD=\angle XBA$ because $A,X,L$ are collinear and $B,X,K$ are collinear.