# Relation between incentre of a triangle and a circle touching its two sides and circumcircle

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

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,

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?

• Must $TP$ and $CM$ intersect on the circle as shown in the diagram, but isn't stated in the writeup? Dec 2, 2021 at 15:44
• @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)
– user997661
Dec 2, 2021 at 15:53
• It would be helpful to clarify that in the writeup (Esp since $R$ is missing from the diagram.) Dec 2, 2021 at 15:54

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 $$U$$.

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!

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