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

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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:

Answer 4

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

$$CN-CA=CB=CM$$

$$AN=MB$$