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The incircle of $\triangle ABC$ is tangent to $AB$, $BC$, and $CA$ at $C'$, $A'$, and $B'$, respectively. Prove that the perpendiculars from the midpoints of $A'B'$, $B'C'$, and $C'A'$ to $AB$, $BC$, and $CA$, respectively, are concurrent.

I have the midpoints of $A'B'$, $B'C'$, and $C'A'$ as $C''$, $A''$, and $B''$. I know there is a homothety relating $\triangle A'B'C'$ and $\triangle A''B''C''$, however, I don't know how to use it. I also do not know what $\triangle A''B''C''$ has to do with the sides of $\triangle ABC$. Can I have some help as to how to prove this geometrically (without any algebra)? Much appreciated.

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2 Answers 2

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Let $\varphi$ the homothety that maps $A',B',C'$ to $A'',B'',C''$, respectively, and let $I$ be the incenter of the triangle $ABC$.

The lines $A'I$, $B'I$ and $C'I$ pass through $A',B',C'$, and they are perpendicular to $BC,CA,AB$, respectively.

The lines $\varphi(A'I),\varphi(B'I),\varphi(C'I)$ are the lines in the problem statement: they pass through $\varphi(A')=A''$, $\varphi(B')=B''$ and $\varphi(C')=C''$, and they are parallel to $A'I, B'I, C'I$, therefore perpendicular to $BC,CA,AB$, respectively. They all pass through the point $\varphi(I)$.

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Here what you have to know: If for two triangle $PQR$ and $P'Q'R'$ the perpendiculars from the points $P$, $Q$, $R$ to the segments $Q'R'$, $R'P'$ and respectively $P'Q'$ are concurrent the also the he perpendiculars from the points $P'$, $Q'$, $R'$ to the segments $QR$, $RP$ and respectively $PQ$ are concurrent. Hence the relation between $PQR$ and $P'Q'R'$ is symmetric.

One shows this by checking that both the properties are equivalent to the equality

$$PQ'^2 +QR'^2 + RP'^2 = P'Q^2 + Q'R^2 + R'P$$

This above equivalence is proved using the following fact: for points $S$ on a perpendicular line to a segment $UV$ the difference $SU^2 - SV^2$ is constant. This follows right away from Pythagoras.

With this result, it's enough to show that the perpendiculars from $A$, $B$, $C$ to the segments $B"C"$, $C" A"$ and $A"B"$ are concurrent. Notice now that $B"C" \parallel B'C'$ and the other, so we have to check the concurrency of the perpediculars from $A$, $B$, $C$ to $B'C'$, $C'A'$ and $A'B'$. But these are the bisectors of the triangle $ABC$. ...

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  • $\begingroup$ Sorry, but can you elaborate a little? I don't quite understand how the two scenarios are related.. $\endgroup$ Commented Jan 14, 2015 at 1:23

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