# How to show that two certain chords of a circle passing through the incentre of a given triangle are equal?

Let $$I$$ be the incenter of a triangle $$\triangle ABC$$. The circle $$AIB$$ meets the sides $$BC$$ and $$AC$$ at points $$M$$ and $$N$$, respectively. I'm trying to prove then that $$BM=AN$$. Here's a figure for reference: Let $$O$$ be the center of the circle $$AIB$$. Obviously, I first tried showing that the triangles $$\triangle BOM$$ and $$\triangle NOA$$ are congruent to each other. In this direction, it's obvious that $$OB=OM=ON=OA$$. But I have no clue as to how to show $$\angle BOM=\angle NOA$$ !!

As that didn't work out, I started considering the cyclic quadrilateral $$\square ANBM$$. I am aware of many interesting properties of cyclic quadrilaterals (Ptolemy's Theorem, opposite angles being supplementary, diagonal intersection, etc.) but none of them seem to help towards actually being able to show that $$BM=AN$$.

Does anyone have any useful comments or hints for this? TIA.

Alternatively, Let $$\angle A = 2a$$, $$\angle B = 2b$$ with (wlog) $$a \ge b$$.

Then $$\angle ABI= b$$ ($$BI$$ is the bisector of $$\angle ABC$$), so $$\angle ANI = b$$ (angles subtended at circumference by same arc are equal). Also, $$\angle IAC = a$$ ($$AI$$ is the bisector of $$\angle BAC$$), so $$\angle AIN = a - b$$ (external angle of triangle is sum of interior opposite angles).

Similarly, $$\angle MBI= b$$ ($$BI$$ is the bisector of $$\angle ABC$$), so $$\angle MAI = b$$ (angles subtended at circumference by same arc are equal). Also, $$\angle IAB = a$$ ($$AI$$ is the bisector of $$\angle BAC$$), so $$\angle BAM = a - b$$ (subtraction).

Hence chords $$AN$$, $$BM$$ subtend equal angles at the circumference of the same circle, so are equal.

• Can you please elaborate on how to show $\angle NIA = b-a = \angle BAM$ ?? Sep 30 at 6:49
• I've done that (and swapped b > a to a > b, to correspond to your diagram). Is this sufficient for you?
– mcd
Sep 30 at 8:22 In the figure, let $$\angle A=2 \alpha, \angle C = 2 \gamma$$ and $$\angle CMI=\beta$$.

Then $$\beta = \alpha$$ (by circle property)

Consequently $$\Delta CAI \cong \Delta CMI$$ (AAS)

$$\therefore CA=CM \tag{1}$$

From circle property, $$CA \times CN = CM \times CB \tag{2}$$

$$(1)$$ and $$(2) \implies$$ $$CN=CB$$

$$CN-CA=CB=CM$$

$$AN=MB$$

• Can you explain what circle property tells us that $\beta=\alpha$ ? Sep 30 at 20:24
• @math-physicist $\beta=\alpha$ i.e. $\angle IMC=\angle IAB$ is true because for a cyclic quadrilateral, exterior angle equals interior opposite angle. Oct 1 at 0:03

Hint: Can you see why $$M$$ is the image of the point $$A$$ reflected in line $$CI$$ and similarly $$N$$ is the image of $$B$$?

• tbh, I have "observed" this while trying to show $BM=AN$ but I don't know of a rigorous reason why. Could you elaborate more? Does this have to do with showing $O,I,C$ are collinear? How does this help in showing $BM=AN$? Sep 30 at 6:06
• Yes, show that $OIC$ is collinear, or equivalently, when $CI$ extended meet the circle again at $D$ then $ID$ is a diameter of the circumcircle. That part is just angle-chase to show $\angle IAD=90^\circ$. So $BM$ is the reflection of $AN$ hence the same length. Sep 30 at 6:14
• ok so couple of things: (1) when you said $ID$ is the diameter of the circumcircle, did you mean the circle $AIB$ (that's not the circumcircle)? (2) after i show that $ID$ is a diameter of circle $AIB$, what allows us to use the reflection argument? I know it's true but I was wondering if there's a theorem I can cite for that? Sep 30 at 6:26
• (1) Yes I mean the circumcircle of $AIB$. (2) Reflect everything in line $CI$. The lines $CA\leftrightarrow CB$ but the circumcircle $AIB$ maps back to itself, so intersecting gives $(A,N)\leftrightarrow (M,B)$. Sep 30 at 6:59
• @math-physicist $\angle BAD=\angle BID=\angle IBC+\angle ICB=\frac12(\angle B+\angle C)$ and $\angle BAI=\frac12\angle A$. So $\angle IAD=\angle BAD+\angle BAI=\frac12(\angle A+\angle B+\angle C)=90^\circ$. Sep 30 at 7:51 We use this property that circle AIB passes outer center P (intersection of bisectors of exterior angles and angle ACB) such that CI is always perpendicular on AN or BM, hence $$BM=AN$$ if triangle ABC is isosceles, as can be seen in figure bellow: 