Problem 5. Let ABC be a triangle with \BCA = 90, and let D be the foot of the altitude from
C. Let X be a point in the interior of the segment CD. Let K be the point on the segment AX
such that BK = BC. Similarly, let L be the point on the segment BX such that AL = AC. Let M
be the point of intersection of AL and BK.
Show that MK = ML.
Here's my drawing but I'm not sure if it's right. Also it's a little clumsy.
I'm having a problem understanding ''D be the foot of the altitude from C''. Am I to draw a angle from C to form D like I do? Because if so, then I find it strange that they word the next paragraph: ''Let X be a point in the interior of the segment CD.'' That doesn't seem to make much sense to call it 'interior' of the segment. It's just segment CD...