euclidean-geometryplane-geometry
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…
$\triangle ABC$ is isosceles. Therefore,
$\measuredangle BAC = \measuredangle C \implies 2\alpha+\alpha =120^\circ\implies \alpha =40^\circ$
Therefore $\measuredangle ABC = 180^\circ - 160^\circ = 20^\circ$
Draw $AX$ such that $\measuredangle XAK = 30^\circ$.
Draw $TX$.
$\measuredangle CDB = 120^\circ ~~\therefore \measuredangle DIA = 90^\circ$
Since $CI$ is angle bisector $IA = IX$
$\triangle ITA \cong \triangle IXT ~(L.A.L) \rightarrow \measuredangle AXT = 60^\circ$
Therefore $\triangle ATX$ is equilateral.
$\measuredangle KXB = 180 ^\circ -110^\circ= 70^\circ$
$AK$ is angle bisector and the height of triangle $ATX$
$ \therefore AKT = 90^\circ$
$TK = KX \rightarrow \triangle KTB \cong \triangle KXB~ (L.A.L)$
Therefore $\triangle TBX$ is isosceles.
$\measuredangle B = 180^\circ -140^\circ =40^\circ$
$\therefore \boxed{\color{red}x=40^\circ-20^\circ=20^\circ}$
Let $K\in AD$ such that $ABCK$ be a parallelogram.
Thus, by law of sines for $\Delta KCD$ we obtain: $$\frac{m}{\sin\theta}=\frac{\frac{11m}{5}}{\sin3\theta}.$$ Can you end it now?
I got $\theta=\arcsin\frac{1}{\sqrt5}.$
Best Answer
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 $\triangle PQM$, $\angle IMN=(135^\circ-\frac{\theta}{2})-(135^\circ-\theta)=\frac{\theta}{2}$