ahmedittihad wrote:**Problem 11**:

Let $ABC$ be a triangle inscribed circle $(O)$, orthocenter $H$. $E,F$ lie on $(O)$ such that $EF\parallel BC$. $D$ is midpoint of $HE$. The line passing though $O$ and parallel to $AF$ cuts $AB$ at $G$. Prove that $DG\perp DC$.

**Solution of problem 11 :**
We first denote some extra points. $H_A,H_B,H_C$ be three projections from $A,B,C$ on $BC,CA,AB$ resp. $K$ be midpoint of $AH$ , & $OG \cap AC = T$

**Claim 1**: $D$ lies on nine-point circle of $\triangle ABC$

**Proof**:

$D , K$ are midpoints of $EH,AH$ resp. Easily can be seen, By $\frac{1}{2}$ scale factor homothety of center $'O'$, $C$ sends to $D$ that lies on $\bigodot KH_AH_BH_C$, called nine point circle of $\triangle ABC$.

**Claim 2**:$\angle BAF = \angle CAE$

**proof:**
$BC\parallel EF \Rightarrow \widehat{BE}=\widehat{CF} \Longrightarrow \angle BAE =\angle CAF$.

Now, by adding $\angle BAC$ to bothsides, we get desired claim.

$\bigstar$ We extend $KD$ both sides that intersects $AB,AC$ at $L$ & $M$ resp.As, $D , K$ are midpoints of EH,AH resp. so, $DK||AE \Rightarrow LM||AE$.

**Claim 3** : $LMTG$ cyclic

** Proof** :

\begin{align*}

\angle LGT&=\angle AGO$\\

&=180^{\circ}- \angle GAF \, ;[\text{as }OG||AF] \\

&= 180^{\circ}- \angle BAF\\

&=180^{\circ}- \angle CAE \, ;[\text{by claim }2]\\

&=180^{\circ}-\angle MAE\\

&= \angle AML\\

\Longrightarrow LMTG & \text{is cyclic.}

\end{align*}

** claim 4**:$CH_CDM$ is cyclic

**Proof**:

[N.B. Here, we use directed angles.]

\begin{align*}

\measuredangle H_CDM&=\measuredangle H_CDK\\

&=\measuredangle H_CH_AK\\

&=\measuredangle H_CCA\\

&=\measuredangle H_CCM\, ;[\text{as }H_ACAH_C]\\

\Longrightarrow CH_CDM & \text {is cyclic.}

\end{align*}

** Claim 5** : $CH_CGM$ is cyclic

**Proof**:

Use

** claim 3 **& Angle chasing. Left for the reader.

By combining

**Claim 4** &

**claim 5**, we get $CH_CGD$ is cyclic

Thus, $\angle CH_CG=90^{\circ} \Longrightarrow \angle CDG =90^{\circ} \Longrightarrow DG \perp DC $ $\blacksquare$