Well, I think the diagram has some problems, I will attach a new one in case anyone needs it. Assume that the radii of the incircles of the two triangles be $R$ and $R'$ respectively, and their difference in area is $S$. Also let $r$ denote the raduis of the six smaller circles, then we have $$S=(a+c+e-b-d-f)r=[(a+c+e+k+l+...+q)r+(k+l+...+q)R]-[(b+d+f+k+l+...+q)r+(k+l+...+q)R']$$ Which simplifies to $R=R'$ as desired.

qinghong wrote: Well, I think the diagram has some problems, I will attach a new one in case anyone needs it. Assume that the radii of the incircles of the two triangles be $R$ and $R'$ respectively, and their difference in area is $S$. Also let $r$ denote the raduis of the six smaller circles, then we have $$S=(a+c+e-b-d-f)r=[(a+c+e+k+l+...+q)r+(k+l+...+q)R]-[(b+d+f+k+l+...+q)r+(k+l+...+q)R']$$ Which simplifies to $R=R'$ as desired. Can you show how to construct the figure?

Observe that mikestro wrote: If this hexagon is removed..., six small triangles remain. is not true since you end up with this kind of object.

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nabodorbuco2 wrote: Observe that mikestro wrote: If this hexagon is removed..., six small triangles remain. is not true since you end up with this kind of object. not the lines, but the interior as well.