Perpendicular lines through the foot of an altitude

For discussing Olympiad level Geometry Problems
User avatar
Phlembac Adib Hasan
Posts:1016
Joined:Tue Nov 22, 2011 7:49 pm
Location:127.0.0.1
Contact:
Perpendicular lines through the foot of an altitude

Unread post by Phlembac Adib Hasan » Mon Oct 10, 2016 9:59 am

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.
Welcome to BdMO Online Forum. Check out Forum Guides & Rules

User avatar
Raiyan Jamil
Posts:138
Joined:Fri Mar 29, 2013 3:49 pm

Re: Perpendicular lines through the foot of an altitude

Unread post by Raiyan Jamil » Mon Oct 10, 2016 11:52 am

Solution:
Here, $\textit{Butterfly Theorem}$ gives us $DE=DF$. Again it is a well known fact that the reflection of $H$, be $G$ lies on the circumcircle of $ABC$. Now, in quadrilateral $HEGF$, the diagonals bisect each other. Therefore, it is a parallelogram. Now, $\angle EHD=\angle AGC=\angle ABC$ which completes the proof.
Math.png
Math.png (153.36KiB)Viewed 4128 times

User avatar
asif e elahi
Posts:185
Joined:Mon Aug 05, 2013 12:36 pm
Location:Sylhet,Bangladesh

Re: Perpendicular lines through the foot of an altitude

Unread post by asif e elahi » Thu Oct 13, 2016 4:10 pm

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
geo4.png
geo4.png (28.89KiB)Viewed 4104 times
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*}

Post Reply