Prove that the triangle formed by the points of contact of the sides of a given triangle with the excircles corresponding to these sides is equivalent to the triangle formed by the points of contact of the sides of the triangle with the inscribed circle
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1what does equivalent mean? – i. m. soloveichik Jul 31 '14 at 13:11
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I think that it means the triangles are congurent – Apoorv Jain Jul 31 '14 at 13:15
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2You don't even know what it means? At least in this problem, the two triangles are not congruent, they have the same area. – Quang Hoang Jul 31 '14 at 13:17
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I know and that's what I am telling you – Apoorv Jain Jul 31 '14 at 13:18
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Moreover I saw this problem in the book and couldn't get it spending hours on it and thus asked you all – Apoorv Jain Jul 31 '14 at 13:19
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At least you should have sketch the picture to check if the triangles are congruent or not, no? – Quang Hoang Jul 31 '14 at 13:22
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I have a property related to it . Orthocentre of excentral triangle is incentre of smaller triangle referred to as pedal triangle . It is useful to solve many problems. – Archis Welankar Oct 28 '15 at 08:02
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Hint: To show the two triangles are equivalent (or having the same area), note that $$area(\triangle AEF)=\frac12(p-a)^2\sin A =\frac 12(p-a)^2\frac{a}{2R} = \frac{(p-a)^2(p-b+p-c)}{4R}$$ where $a,b,c$ are the sides of $\triangle ABC$ and $p=\frac{a+b+c}{2}$.
Similarly for the triangles $\triangle AE'E',\dots$
Quang Hoang
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