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:

enter image description here

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?

  • $\begingroup$ Hint: $AK = AP = \frac12(b+c-a)$ and similar relations for $AM$ and $ND$. $\endgroup$ Nov 15, 2018 at 23:47
  • $\begingroup$ Couldn't do it, still. Feeling a little bad about it, it doesn't seem to be a hard problem.... $\endgroup$ Nov 16, 2018 at 12:23
  • $\begingroup$ 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... $\endgroup$ Nov 16, 2018 at 12:25
  • $\begingroup$ Yeah, I guess that's the opened form of $p - a$, isn't it? Tried that. But can't see how that helps. $\endgroup$ Nov 16, 2018 at 12:28

1 Answer 1


Before we start, let us recall how to compute the distance between a vertex and the nearby points where the incircle touch the triangle.

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.


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