I have the following question with me:
"Altitudes $AD$ and $CE$ of an acute angled triangle $ABC$ intersect at point $H$ . Let $K$ be the midpoint of side $AC$ and $P$ be the midpoint of segment $DE$ . Let $Q$ be the symmetrical point of $K$ wrt line $AD$. Prove that $\angle QPH = 90^o$"
So basically I need to prove that $QP$ and $PH$ are perpendicular to each other.
My observations:
If $F$ is the point of intersection of $AD$ and $QK$, then I have $PF$ parallel to $AE$ by the mid-point theorem and hence I have angle $EPF$ equal to angle $C$. However I was unable to proceed further.
What the solution says:
The solution booklet tells me to take the reflection of $Q$ across $P$ to give $R$ and then I figured out that KR would be equal to and parallel to $AE$, however the solution further states that $R$ is the reflection of $K$ in $CE$. how did he conclude this?
Also can anyone help me finish the proof that I started above?
Best Answer
I will use your notation, but also denote by $M$ the intersection of $KR$ and $CE$.
As you noted, $FP$ is a midline in $\triangle DAE$, so $FP\parallel AE$ and $FP=\frac{1}{2}AE$.
$FP$ is a midline in $\triangle QKR$ as well, so $FP\parallel KR$ and $FP=\frac{1}{2}KR$.
By 1. and 2. $AE\parallel KR$ and $AE=KR$.
In $\triangle CAE$, $KM\parallel AE$ and $K$ is a midpoint of $CA$, so $KM$ is a midline in this triangle, implying $KM=\frac{1}{2}AE$.
By 3. and 4. $KM=MR$ and $CE\perp KR$ (because $CE\perp AE\parallel KR$), so $R$ is the reflection of $K$ with respect to $CE$.
Now note: $HQ=HK=HR$, where the first equality holds since $H$ lies on the perpendicular bisector $AD$ of $QK$, and the second one since $H$ lies on the perpendicular bisector $CE$ of $KR$.
But $HQ=HR$ implies that $H$ lies on the perpendicular bisector of $QR$, and since this line passes through $P$ we get $\angle QPH=90^\circ$.