MathLabs

Problem 5

Let ABCABC be a triangle with ∠BCA=90∘\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=BCBK=BC, and L lie on BX with AL=ACAL=AC. Let M=AL∩BKM=AL\cap BK. Prove MK=MLMK=ML.
Step 2 of 5: Introduce the auxiliary point R
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In plain words

A point R beyond C is chosen so two right triangles become similar and the later circle appears naturally.

DX⋅DR=BD⋅ADDX\cdot DR=BD\cdot AD
Detailed analysis

Choose R on the ray from D through C beyond C so that DX⋅DR=BD⋅ADDX\cdot DR=BD\cdot AD. Since ∠BDX=∠RDA=90∘\angle BDX=\angle RDA=90^\circ, triangles RADRAD and BXDBXD are similar. Hence ∠XBD=∠ARD\angle XBD=\angle ARD, and step 1 gives ∠ALD=∠ARD\angle ALD=\angle ARD.