Prove that AM is perpendicular to Bh

Question

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 $AM$.

Answer 1

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

Answer 2

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 $AQ=AM$.

Answer 3

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 $(\vec{HD}, \vec{AH} + \vec{BH})=0$