In △ABC (not isosceles) CH, CL and CM are respectively height, bisector and median. Show that ∠ACB=90 degrees if and only if ∠HCL=∠MCL.
I think that I have to show that △MCB (or △ACM) is isosceles, but I can't figure it out. I will be very grateful if you help me.
Search found 1 match
Search found 1 match • Page 1 of 1