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Let $\triangle ABC$ be a scalene triangle with point $D$ inside it. Let $AD$, $BD$ and $CD$ intersect $BC$, $CA$ and $AB$ in $M$, $N$ and $O$. Let the midpoint of the segment that connects the incenters of the triangles $\triangle BDO$ and $\triangle CDN$ be $D1$. Similarly define points $D2$ and $D3$.
It is true: $AN + BO + CM = AO + CN + BM$.

Prove that $D, D1, D2, D3$ are on the same circle using Ptolemy's trigonometric theorem.

I tried to use Ptolemy for $D, D1, D2, D3$ points but the angles were not right. I computed the distance between $D$ and incenters. I tried to compute the angles between $D1$ and $D$, and all the others but they do not seem to get simplified.
I used Ceva for $D$.
I tried to prove that $D$ is the incenter of $\triangle ABC$.
But anything that I used was too complicated to compute.

Can you find examples of Point $D$?

Here is a picture. Can $I1$, $I2$ and $D$ be colinear? I could not prove it, but in Geogebra it seems so.

CONCYCLIC POOINTS

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    $\begingroup$ Can you make a drawing of this? It sounds quite complicated 😊 $\endgroup$ Commented Jan 16 at 10:46
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    $\begingroup$ 1) How is P defined ? 2) What do you mean by "It is true : etc..." : is it a fact valid for any triangle $ABC$ or do you assume thet we are in this case ? $\endgroup$ Commented Jan 16 at 11:04
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    $\begingroup$ @Dominique: picture added. Thanks for the suggestion. $\endgroup$ Commented Jan 16 at 14:58
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    $\begingroup$ Instead of "the center of the circle inscribed in the triangle $\triangle BDO$" you could write the clearer "the incenter of $\triangle BDO$" $\endgroup$ Commented Jan 16 at 15:02
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    $\begingroup$ In connection with my previous remark , I just asked a connected question here $\endgroup$ Commented Jan 17 at 19:18

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