What´s the measure of angle $\angle IMN$?

Question

For reference:

In a triangle $ABC$, $\hat B=90^\circ$, $AB < BC$, draw the height $BH$ where $M$, $N$ and $I$ are the incenters of $ABH$, $HBC$ and $ABC$.

$\measuredangle BCA=\theta$.

Calculate $\measuredangle IMN$.

My progress ... Here's a drawing with the relationships I found...

Answer 1

Let $BH$ intersect $MI$ and $MN$ at $P$ and $Q$ respectively.

In two similar triangles, ratio of corresponding sides=ratio of inradii

$\small{\triangle ABH\sim\triangle BCH}\implies\frac{AB}{BC}=\frac{GH}{HF}=\frac{MH}{HN}\implies\small{\triangle ABC\sim\triangle MHN}$

Therefore, $\angle HMN=90^\circ-\theta$

$\angle MQP=45^\circ+(90^\circ-\theta)=135^\circ-\theta$

$\angle ABH=\theta, \angle BAD=45^\circ-\frac{\theta}{2}\implies\angle BPM=135^\circ-\frac{\theta}{2}$

Therefore from $\triangle PQM$, $\angle IMN=(135^\circ-\frac{\theta}{2})-(135^\circ-\theta)=\frac{\theta}{2}$

Answer 2

Hint: Start with a simple form where triangle is isosceles. Observe that points A, M and I are collinear. Similarly points C, N and I are collinear.In this case $\angle IMN=\angle INM= \frac{\angle BCA}2$. You have to show that $\angle IMN=\frac {\theta}2$ in all positions of vertex B on the circumcircle. Also $\angle MIN=135^o$ all time.

Answer 3

Extend $CN$ and it intersects $BM$ at $J$. Extend $BM$ and it intersects $AC$ at $K$.

Given $\angle ABH = \angle ACB = \theta$ and $BK$ bisects $\angle ABH$ so we have,

$CJ \perp BK$ but $BJ$ is also the angle bisector of $\angle BCK$. Hence $J$ is midpoint of $BK$ and circumcenter of right triangle $\triangle BHK$

So, $\angle BHJ = \angle HBJ = \theta/2$

As $JMHN$ is cyclic, $\angle JMN = \angle JHN = 45^\circ + \theta/2$

As $\angle AMB = 135^\circ, \angle IMJ = 45^\circ$

$\therefore \angle IMN = \theta/2$

Answer 4

Yet an other solution chasing angles. Let $2\alpha$ and $2\gamma=\theta$ be the angles in $A$ and respectively $C$, so $90^\circ=2(\alpha+\gamma)$ giving $\alpha+\gamma=45^\circ$.

We draw now the angle bisectors in the mentioned three triangles from the posted problem, and have the picture:

Construct the intersections $M'$ of the angle bisectors $AM$ and $BN$, and $N'$ of the angle bisectors $CN$ and $BM$.

We have $CNN'\perp BN'M$ since in $\Delta CN'B$ we know the angles in $B,C$. So $NN'$ is a height in $\Delta BMN$. Similarly $MM'$ is also a height. So we have in one breath: $$ \widehat{IMN}= \widehat{IBN}=\gamma =\frac 12\theta\ . $$