# Hard geometry question (circles and triangles)

Could anyone solve this one for me? I tried to use some kind of variation on Poncelet's theorem, but couldn't find a way to do it. Here it goes:

Let $$ABC$$ be a triangle and $$D\in BC$$ such that $$AD\perp BC$$. Also, let $$I$$ and $$J$$ be the incenters of triangles $$ABD$$ and $$ACD$$, respectively. The incircles of $$ABD$$ and $$ACD$$ are tangents to $$AD$$ on dots $$M$$ and $$N$$, respectively. Finally, let $$P$$ be the point of tangency of $$ABC$$'s incircle with side $$AB$$. The circle centered in $$A$$ and with radius $$AP$$ intersects $$AD$$ in $$K$$.

a) Show that triangles $$IMK$$ and $$KNJ$$ are congruent.

b) Show that $$IDJK$$ is cyclical.

I didn't have a problem with the image. Here's how I saw it:

I couldn't find a relation between the last circle built and the other ones. And to prove congruence I have to do so, since $$K$$ is one of both triangles' vertices, and it's built from that specifical circle. So, anyone, please?

• Hint: $AK = AP = \frac12(b+c-a)$ and similar relations for $AM$ and $ND$. – achille hui Nov 15 '18 at 23:47
• Couldn't do it, still. Feeling a little bad about it, it doesn't seem to be a hard problem.... – Italo Marinho Nov 16 '18 at 12:23
• Did you get what the hint tell you? i.e. the distance between the contact point and the vertex can be computed from the side lengths... – achille hui Nov 16 '18 at 12:25
• Yeah, I guess that's the opened form of $p - a$, isn't it? Tried that. But can't see how that helps. – Italo Marinho Nov 16 '18 at 12:28

Given any triangle $$ABC$$, let $$a = BC, b = CA, c = AB$$. The incircle of $$ABC$$ will touch the sides of triangle at three points. Let $$P, Q, R$$ be the points on sides $$AB$$, $$BC$$, $$CA$$ respectively. Notice $$\begin{cases}AP = AR,\\ BP = BQ,\\ CQ = CR\end{cases} \quad\text{ and }\quad \begin{cases} AP + PB = AB = c,\\ BQ + QC = BC = a,\\ CR + RA = CA = b \end{cases}$$ We have \begin{align}AP &= \frac12(AP + AR)\\ &= \frac12((AP+PB) + (AR+RC) - (QC+BC))\\ &= \frac12(b + c - a)\end{align}
For the problem at hand, let $$h = AD, u = BD, v = CD$$, we have $$a = u+v$$.
Apply above theorem to $$\triangle ABC$$, $$\triangle ADB$$ and $$\triangle DAC$$, we get $$AK = AP = \frac12(b+c-a),\quad AM = \frac12(h+c-u)\quad\text{ and }\quad ND = \frac12(h+v-b)$$ This leads to \begin{align}KM = AM - AK &= \frac12( (h + c - u ) - (b+c - (u+v))\\ &= \frac12(h + v - b) = ND\end{align} Since $$\angle ADC = 90^\circ$$, $$ND = JN$$ and hence $$KM = JN$$. Similarly, \begin{align}KN = AD - AK - ND &= h - \frac12(b+c-a) - \frac12(h+v-b) \\ &= \frac12(2h - b - c + (u+v) - h - v + b)\\ &=\frac12(h + u - c) = MD\end{align} Since $$\angle BDA = 90^\circ$$, $$MD = IM$$ and hence $$KN = IM$$.
It is clear $$\angle IMK = \angle KNJ = 90^\circ$$. By SAS, $$\triangle IMK$$ is congurent to $$\triangle KNJ$$. Notice \begin{align}\angle DJK + \angle KID &= ( \angle DJN + \angle NJK ) + ( \angle KIM + \angle MID )\\ & = (45^\circ + \angle NJK ) + (\angle JKN + 45^\circ)\\ &= 90^\circ + (\angle NJK + \angle JKN)\\ &= 90^\circ + (180^\circ - \angle KNJ)\\ &= 90^\circ + 180^\circ - 90^\circ = 180^\circ\end{align} The quadrilateral $$IKJD$$ is cyclic.