# Thread: Circle Geometry: Incenter perpendicular to Chord?

1. ## Circle Geometry: Incenter perpendicular to Chord?

Bisector of the angles A,B,C of a triangle ABC meet the circumcircle of the triangle in points P, Q and R respectively and intersect at the point T. Prove that Ap is perpendicular to RQ, BQ is the perpendicular to RP and CR is the perpendicular to QP.

I tried to prove that T is the centre of the circle. Then i would be perpendicular to chord. But i have no clue

2. Originally Posted by Lukybear
Bisector of the angles A,B,C of a triangle ABC meet the circumcircle of the triangle in points P, Q and R respectively and intersect at the point T. Prove that Ap is perpendicular to RQ, BQ is the perpendicular to RP and CR is the perpendicular to QP.

I tried to prove that T is the centre of the circle. Then i would be perpendicular to chord. But i have no clue
That's because T doesn't necessarily have to be the center of the circumcircle. However, it is the incenter of the triangle. (the incenter is the intersection of the internal angle bisectors). It's been a while since I took my transformation geometry course, but I would probably begin with an accurate drawing using a compass and straight edge.

My intuition tells me that you need to perform a dilation transformation with center T with factor -TQ/TB. then something follows from there.

3. Hello, Lukybear!

I assume you have a good sketch.

Bisectors of the angles $A,B,C$ of $\Delta ABC$ meet the circumcircle
of the triangle in points $P, Q, R$ respectively and intersect at point $T.$

Prove that: . $AP \perp RQ,\; BQ \perp RP,\;CR \perp QP.$

Let the angles at $A$ be $\alpha.$
That is: . $\angle BAP = \angle CAP = \alpha.$
Then: . $\text{arc}(BP) = \text{arc}(PC) = 2\alpha$

Let the angles at $B$ be $\beta.$
That is: . $\angle ABQ = \angle CBQ = \beta.$
Then: . $\text{arc}(AQ) = \text{arc}(QC) = 2\beta$

Let the angles at $C$ be $\gamma.$
that is: . $\angle ACR = \angle BCR = \gamma.$
Then: . $\text{arc}(AR) = \text{arc}(RB) = 2\gamma$

In $\Delta ABC\!:\;2\alpha + 2\beta + 2\gamma \:=\:180^o \quad\Rightarrow\quad \alpha + \beta + \gamma \:=\:90^o$

Let $AP$ and $RQ$ intersect at $U.$

From the "intersecting chords theorem",
. . we have: . $\angle AU\!R \;=\;\dfrac{\text{arc}(AR) + \text{arc}(QP)}{2}$

Hence: . $\angle AU\!R \:=\:\dfrac{2\gamma + (2\beta + 2\alpha)}{2} \;=\;\alpha + \beta + \gamma \:=\:90^o$

Therefore: . $AP \perp RQ.$

In a similar fashion, prove that $BQ \perp RP$ and $CR \perp QP.$