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Thread: Circle Geometry: Incenter perpendicular to Chord?

  1. #1
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    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
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  2. #2
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    Quote Originally Posted by Lukybear View Post
    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.
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  3. #3
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    Hello, Lukybear!

    I assume you have a good sketch.


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

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

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

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

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

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


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

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

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

    Therefore: .$\displaystyle AP \perp RQ.$


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

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