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In the figure, $P$, $Q$ and $T$ are points of tangency. Prove that $M$ is the incentre of $\triangle ABC$.

book_image

Here's what I did so far:

[Note: According to the diagram, $M$ is the intersection of lines $PQ$ and $CR$. (See the figure below)]

Extend $TQ$. Then,

IMG


  1. $\rlap{\underbrace{\phantom{\angle PQT=\angle RST}}_{\because\: PQ\parallel RS}}\angle PQT=\overbrace{\angle RST=\angle RCT}^{\text{angles of same segment}}\\ \implies CQMT \text{ is cyclic.}$

  1. $\rlap{\underbrace{\phantom{\angle QPT=\angle CQT}}_{\text{alternate segment theorem}}}\angle QPT=\overbrace{\angle CQT=\angle CMT}^{\text{angles of same segment}}\\ \implies\triangle MPR\sim\triangle TMR$

It is sufficient to show that points $A$, $M$ and $B$ lie on a circle centered at $R$. i.e. $RA=RM=RB$. Any ideas to proceed?

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  • $\begingroup$ Must $TP$ and $CM$ intersect on the circle as shown in the diagram, but isn't stated in the writeup? $\endgroup$
    – Calvin Lin
    Dec 2, 2021 at 15:44
  • $\begingroup$ @CalvinLin ... yes, we are supposed to use the data given through the diagram. (And point M is defined as the intersection of PQ and CR) $\endgroup$
    – user997661
    Dec 2, 2021 at 15:53
  • $\begingroup$ It would be helpful to clarify that in the writeup (Esp since $R$ is missing from the diagram.) $\endgroup$
    – Calvin Lin
    Dec 2, 2021 at 15:54

1 Answer 1

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In fact, this is a direct result of Y. Sawayama's lemma:

Through vertex $A$ of $\triangle ABC$ a straight line $AD$ is drawn with $D$ on $BC$. Let circle $C_1$ tangent to $AD$ at $F$, $CD$ at $E$, and the circumcircle $C_2$ of $\triangle ABC$ at $K$ be centered at $P$. Then the chord $EF$ passes through the incenter $I$ of $\triangle ABC$.

Whilst the link above addresses a more general case, I have completed my work using some ideas presented there.


First note that $R$ is the midpoint of $\overset{\huge\frown}{AB}$.

Proof:

Extend $BA$ to meet tangent drawn to circles at $T$, and call the intersection point $U$.

image

Being tangents to inner circle, $UP=UT$ and therefore $\angle TPU=\angle PTU=y+z$.

By alternate segment theorem $\angle ABT=\angle ATU=z$.

From exterior angle property, $\angle UAT=\angle ABT+\angle ATB=\angle APT+\angle ATP\implies z+x+y=2y+z\implies x=y.$

Therefore $TR$ bisects $\angle ATB$ and hence the result.


We have already proved that $\triangle MPR\sim\triangle TMR$. Then, $$\dfrac{RM}{RT}=\dfrac{RP}{RM}\implies RM^2=RP\cdot RT$$

Also $\angle RBA=\angle RCB=\angle RTB$. According to converse of alternate segment theorem, $RB$ is tangent to circumcircle of $\triangle BPT$. Thus we apply tangent-secant theorem to get $RB^2=RP\cdot RT$.

Hence, $RA=RM=RB$ and we are done!

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    $\begingroup$ Indeed it works. $\endgroup$
    – Math Lover
    Dec 3, 2021 at 21:21
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    $\begingroup$ @MathLover .. I found that this special case of Sawayama's lemma is known as Verrier's lemma. (though I could not find many references) $\endgroup$
    – ACB
    Jan 28 at 15:59
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    $\begingroup$ @ACB thanks for letting me know. I never heard of it or may have read and have forgotten about it with age. Will check. $\endgroup$
    – Math Lover
    Jan 28 at 16:03

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