BdMO Online Forum

In an acute triangle $ABC$ with circumcenter $O$ and orthocenter $H$, let $D$ be the foot of the altitude from $A$. The line perpendicular to $OD$ at $D$ meets segment $AB$ at $E$. Prove $\angle DHE=\angle ABC$

Rules: trig bashing is allowed, but not encouraged. Look for synthetic solutions.

Solution:

Let $M$ be the midpoint of $BC$.
Applying Cotangent rule in $\triangle AED$, we get
\begin{align*}
\cot \angle DHE &=\frac{ HD\cdot \cot \angle HAE- AH\cdot \cot \angle HDE}{AD} \\
& = \frac{HD\cdot\frac{AD}{BD}-AH\cdot \cot \angle ODM}{AD}\\
& = \frac{HD}{BD}-\frac{AH\cdot \frac{DM}{OM}}{AD}\\
& = \frac{CD}{AD}-\frac{2DM}{AD}\\
& = \frac{BD}{AD}\\
& = \cot \angle B\\
\Rightarrow \angle DHE &= \angle ABC
\end{align*}
Cotangent rule is a very useful tool when lots of Cevians are involved and you can't do angle chasing easily. Here are the $3$ theorems If $P$ divides $BC$ in the ratio $m:n$ and angles $\theta,\beta$ nd $\gamma$ are shown above, then
\begin{align*}
m\cot C-n\cot B &= (m+n)\cot \theta\\
n\cot \gamma-m\cot \beta &=(m+n)\cot \theta \\
m\cot \beta &=(m+n)\cot A+n\cot B
\end{align*}