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Prove that if the incircle of triangle $ABC$ touches side $BC$ at $D$ and the $A$-excircle touches side $BC$ at $D'$, then the midpoint of $BC$ is the midpoint of $DD'$.

This is an interesting property that I discovered when doing a few problems but the solutions didn't prove it. After drawing several triangles and their in- and excircles, it seems to be true that the midpoint of the intouch and and extouch points is in fact the midpoint of the side of the triangle.

As another question, if anyone can prove that the antipode of the intouch $D$ which I have labeled as $U$ in the triangle below is collinear with the exsimilicenter $A$ and $D'$ that would help. Finally, is it also true that $A,U,D',$ and $I_a$ are collinear or just $A,U,$ and $D'$? enter image description here

Note: $I_a$ is the center of the $A$-exircle.

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  • $\begingroup$ You wrote "incenter." Did you mean "incircle"? Also, "touches" might be better than "intersects" in that case. $\endgroup$ – David Jan 23 '16 at 4:09
  • $\begingroup$ @David Yes, that was a typo. That would be impossible, anyway. $\endgroup$ – Puzzled417 Jan 23 '16 at 4:10
  • $\begingroup$ Are you sure that your definition of $D$ and $D'$ is the same throughout your question and in the illustration? $\endgroup$ – David Jan 23 '16 at 4:24
  • $\begingroup$ @David Yep. An intouch is the point of contact between the incircle and its triangle. $\endgroup$ – Puzzled417 Jan 23 '16 at 4:28
  • $\begingroup$ First you call the intouch $D'$ and later $D$. $\endgroup$ – David Jan 23 '16 at 4:29
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I will answer the first question.

This is easy to prove using just one basic idea: when a circle is tangent to two sides of an angle, the distance from the vertex to each of the points of tangency is the same.

Applying that idea to the incircle, you'll find after some calculations that $BD = \frac{1}{2}(a + c - b)$. Applying it to the excircle opposite vertex $A$, you'll find $CD' = \frac{1}{2}(a + c - b) = BD$.

For the second question, the dilation $h$ with centre $A$ taking $I$ to $I_a$ transforms the incircle into the excircle. It also transforms line $ID$ into a parallel line passing through $I_a$. Since $ID$ and $I_aD'$ are both perpendicular to $BC$, the image of line $ID$ must be line $I_a D'$. Points $U$ and $D$ are the intersections of the incircle with $ID$. They must be carried by $h$ to the intersections of the excircle with $I_a D'$. Since $h$ has positive ratio, $U$ must be carried to $D'$. This proves that $A, U, D'$ are aligned.

Assume $I_a$ is on line $AD'$. Then since $I_a D'$ and $UD$ are both perpendicular to $BC$ and pass through $U$, we must have $D = D'$. This shows that $AI$ is is perpendicular to $BC$, hence that $ABC$ is isosceles.

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  • $\begingroup$ I got it thanks! $\endgroup$ – Puzzled417 Jan 23 '16 at 4:34
  • $\begingroup$ Can you try the second part? $\endgroup$ – Puzzled417 Jan 23 '16 at 4:36
  • $\begingroup$ I don't get the last sentence: how does the proof make it apparent that $I_a$ is on line $AD'$ if and only if $ABC$ is isosceles? $\endgroup$ – Puzzled417 Jan 23 '16 at 11:42
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Finally, is it also true that A, U, D’, and Ia are collinear, or just A, U, and D’ ?

Ia is not collinear with the other three, unless the triangle is isosceles. As an aside, the points A, D, and U’ are also collinear, where U’ is the antipode of D’ with regard to Ia. Also, $DU\parallel D'U'.$

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  • $\begingroup$ If we let $P$ be the point on line $AI$ for which $MD = MP$, is it true that $DPQD'$ is a cyclic quadrilateral with circumcenter $M$? $\endgroup$ – Puzzled417 Jan 23 '16 at 12:50
  • $\begingroup$ @Puzzled417: What is Q ? $\endgroup$ – Lucian Jan 23 '16 at 12:54
  • $\begingroup$ $Q$ is a point on the incircle such that $\angle{AQD} = 90^{\circ}$. $\endgroup$ – Puzzled417 Jan 23 '16 at 13:00
  • $\begingroup$ @Puzzled417: You can answer this and many other such questions for yourself using GeoGebra. :-$)$ $\endgroup$ – Lucian Jan 23 '16 at 13:02
  • $\begingroup$ Yes, but what is the geometrical reason why? $\endgroup$ – Puzzled417 Jan 23 '16 at 13:05

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