# URGENT Geometry - triangle (angle bisector, median, altitude)

• Oct 19th 2008, 04:33 AM
james_bond
URGENT Geometry - triangle (angle bisector, median, altitude)
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.
• Oct 19th 2008, 05:04 AM
vincisonfire
DB = DC gives us CDB = 180 - 2 * BCD
AD = DB gives us ADC = 180 - 2 * ACD
thus
90 = ACD + BCD
• Oct 19th 2008, 06:21 AM
james_bond
Yeah, I know that but what if $\displaystyle \overline{DC}\ne r$. Sorry for the misleading figure - $\displaystyle C$ is not on the circle (if it were then the triangle would be right-angled - Thales' theorem).
• Oct 19th 2008, 06:25 AM
vincisonfire
Yes but $\displaystyle \overline{AD}\ne \overline{BD}$. I mean the circumcenter won't be $\displaystyle D$, the midpoint of $\displaystyle \overline{AB}$.