AD = DB = DC = radius
DB = DC gives us CDB = 180 - 2 * BCD
AD = DB gives us ADC = 180 - 2 * ACD
CDB + ADC = 180
thus
90 = ACD + BCD
Prove that a triangle in which the median and the altitude is symmetric with respect to the angle bisector from the same vertex must have a right angle in this vertex. (Please take a look at the image attached.)
I can prove that if the triangle is right-angled, then the symmetry holds but I can't prove that if it isn't right-angled then they aren't symmetric.