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 5 of 5: Finish with congruent right triangles

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In plain words

The equal hypotenuse-side data make the two right triangles congruent, so the desired segments are equal.

\angle RLM=\angle RKM=90^\circ\Longrightarrow\triangle RLM\cong\triangle RKM\Longrightarrow MK=ML

Detailed analysis

Because L lies on AX and M lies on AL, while K lies on BX and M lies on BK, the perpendiculars above give $\angle RLM=\angle RKM=90^\circ$ . Together with $RL=RK$ and the common hypotenuse RM, triangles $RLM$ and $RKM$ are congruent by hypotenuse-leg, so $MK=ML$ .