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 $ND$. $\endgroup$ – achille hui Nov 15 '18 at 23:47
  • $\begingroup$ Couldn't do it, still. Feeling a little bad about it, it doesn't seem to be a hard problem.... $\endgroup$ – Italo Marinho Nov 16 '18 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... $\endgroup$ – achille hui Nov 16 '18 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. $\endgroup$ – Italo Marinho Nov 16 '18 at 12:28

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.