# Prove that AM is perpendicular to Bh [duplicate]

In the isosceles triangle $$ABC$$, $$M$$ is the median of $$HD$$ and $$AH$$ is perpendicular to $$BC$$ and $$HD$$ is perpendicular to $$AC$$. Prove that $$BD$$ is perpendicular to $$AM$$.

## marked as duplicate by Aqua, Américo Tavares, Jack Lee, Lord Shark the Unknown, Lee David Chung LinApr 27 at 5:42

• Use the Ratio-and-Proportion logic invoked in Similar Triangles. – Sharat V Chandrasekhar Apr 26 at 15:29
• This the third time you've tried asking the same question. If you have concerns or questions about the answers to your initial version, please bring them up there; asking the same question repeatedly trying to get different answers is frowned upon. – Steven Stadnicki Apr 29 at 16:52

Here is an other solution. Consider the heights $$AH$$, $$BB'$$ and $$CC'$$ in the given triangle $$\Delta ABC$$, call $$T$$ their intersection, the orthocenter of $$\Delta ABC$$, and draw $$DD'$$ parallel to $$CC'$$, $$D'$$ on the side $$AB$$.

• Then $$D$$ is the mid point of $$CB'$$, because $$H$$ is the mid point of $$BC$$.

• By the same argument, $$Q$$ is the mid point of $$TB'$$.

• Also, by construction, $$Q$$ is the intersection of two heights of $$\Delta ABD$$, so it is its orthocenter, so $$AQ\perp BD$$. (This is the idea of the proof, use the fact that the third height is perpendicular on the third side.)

• Because $$TB'\|HD$$, the prolongation of $$AQ$$ intersects $$HD$$ in its mid point, which is the point $$M$$ in the problem.

• So $$BD$$ is perpendicular on the line $$AQ=AM$$.

You can draw the point $$K$$ on $$HA$$ such that $$A$$ is the midpoint of $$HK$$ then prove that $$BHDK$$ is cyclic

Use scalar product:

$$(\vec{BD}, \vec{AH} + \vec{AD}) = (\vec{BH} + \vec{HD}, \vec{AH} + \vec{AD}) = (\vec{BH},\vec{AD}) + (\vec{HD}, \vec{AH}) = (\vec{BH}, \vec{HD}) + (\vec{HD}, \vec{AH}) = (\vec{HD}, \vec{AH} + \vec{BH})$$

To show that it is zero, consider the midpoint of AB, let's call it P. Let $$\angle ABC = \angle ACB = \alpha$$. Then $$\angle PHA = \angle PAH = 90^\circ - \alpha$$ and $$\angle AHD = 90^\circ - \angle HAC = 90^\circ - (90^\circ - \angle ACB) = \alpha$$, so $$PH\perp HD$$ and $$(\vec{HD}, \vec{AH} + \vec{BH})=0$$