I have a triangle $\triangle ABC$, with circumcentre $O$, the reflection of $A$ through $BC$ gives me $T$. The perpendicular bisector of $TC$ and $TB$ intersect $AB$ and $AC$ at $P$ and $Q$ respectively.
I need to prove that $TO$ is perpendicular to $PQ$.
How do I prove it? I tried some simple angle chasing but couldn't come to any significant conclusion. The solution booklet however tells me to use the fact that the reflection of $O$ in $BC$, say $O'$ lies on the circumcircle of $\triangle APQ$, how did he arrive at this conclusion?
Best Answer
Let $PR$ be the perpendicular bisector of side $TC$ , and $SQ$ be the perpendicular bisector of side $BT$ . Let them intersect at point $O’$.
Observe that $O’$ is the circumcentre of $\triangle BCT$ . It is thus the reflection of $O$ in $BC$.
$$∠RPA + ∠SQA = ∠O’PA+O’QA = (360-90-2∠C-A)+(360-90-2∠B-A) $$ $$=720-180-2(∠A+∠B+∠C) = 720 - 540 = 180 .$$
Thus , the angles are supplementary. This implies $O’PAQ$ is a cyclic quadrilateral, and the points are concyclic.
Join the diagonals and $OO’$ . Join $O’A$ and $OT$, to meet at $G$. Let $OO’$ meet $BC$ at $D$ . Let $AT$ meet $BC$ at $H$ .
For convenience , Let $∠BAH = w , ∠HAG = y , ∠GAQ = x$ .
Using the properties of cyclic quadrilaterals and by angle chasing , we get $∠BMP = y .$ Also observe that $ ∠HAG = ∠HTG = y$ and since $AT$ is parallel to $OO’ , ∠TOO’ = y.$
Then , the line segments $PQ$ and $OT$ are inclined to $BC$ and $OO’$ respectively at an equal angle . This implies that the rotation of these line segments by that angle , will not change the angle between them , and the angle between them will equal $\angle ODB$. But $∠ODB = 90.$
Hence , OT is perpendicular to PQ