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In $\triangle ABC,$ the incircle touches $AB$ at $V$ and $BC$ at $T.$ $B'$ is the midpoint of $AC$ and $A'$ is the midpoint of $BC.$ $VT$ extended and $B'A'$ extended intersect at $P.$ Prove that $AP$ passes through the incentre, $I,$ of $\triangle ABC.$

I don't know how to go about proving this. I tried using law of sines, bisector theorem, I tried proving that $\angle AIP=180°,$ and that $AP=AI+IP$ but there's always something missing. I'd appreciate any help in the right direction.

Tom
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1 Answers1

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By midpoint theorem, we have A(V)B // P(A’)B’. Then, $\angle 1 = \angle 2 = \angle 3= \angle 4$.

enter image description here

Let (1) the in-circle touches AC at U; (2) BI extended cut PA’B’ at M. From A’ B = A’M = A’C, we can say that M is also on the circle TIUC with IC as the diameter and $\angle BMC = 90^0$.

[Explanation added: A’ B = A’M = A’C Implies A' is the center of the circle that passes though B, M, C. Then, $\angle BMC = 90^0$ by angle in semi-circle.

Now, $\angle IMC = \angle IUC = 90^0$ implies UICM is cyclic. Since UITC must be cyclic by tangent properties, M is the 5th concyclic point circle UITC.]

Note also that CM // PTNV because $\angle BNT = \angle CMX = 90^0$. Then, $\angle 5 = \angle 1 = \angle 4$ implies P is another con-cyclic point of the circle TIUC.

[Explaination added: $\angle 5 = \angle 4$ is a sufficient condition to say TPCM is cyclic.]

[Please ignore:- Since $\angle AIM$ and $\angle MCT$ are equal to 2x + y independently, IA is the extended line of the cyclic quad MCPI.]

[Newly added:-]

enter image description here $\angle (1) = \angle (2)$, (angle at center and angle at circumference)

$ = \angle (3)$, (ext. angle, cyclic quad of the red circle)

$ = \angle (4)$, (angles in the same segment)

$ \alpha = \beta$, (angles in the same segment)

$ = \angle (2)$, (ext. angle cyclic quad.)

That is, $ \alpha = \angle (1)$. And this means IA is the extended part of the cyclic quadrilateral UCPI.

Mick
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  • How do you get from $A'B=A'M=A'C$ to $M$ being on circle $TIUC?$ Also, how does $\angle5=\angle1=\angle4$ imply that $P$ is on the same circle as well? – Tom Jul 09 '19 at 12:57
  • @tom See explanation added. – Mick Jul 09 '19 at 15:05
  • Thanks, this helped me solve the problem. However, I must note a mistake in your drawing $\angle UIA\neq\angle ACB.$ And from that you conclude that $\angle AIM=\angle MCT,$ where in fact $\angle AIM=\angle MCP.$ – Tom Jul 09 '19 at 19:05
  • @tom Good catch. Fixed. See the [Please ignore:-] and the [Newly added:-] – Mick Jul 10 '19 at 04:44