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 4 of 5: Compare the two right triangles

In plain words

Because R lies on the perpendicular through D, its powers relative to the two legs differ by exactly the same amount.

RL^2=RK^2

Detailed analysis

Since $RC\perp AB$ , the right-triangle identities give $AR^2-AC^2=BR^2-BC^2$ . Hence $RL^2=RK^2$ , so $RL=RK$ .