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I read this proof about the construction of the symmedian. It says that if we want to construct the symmedian for a triangle $ABC$ one way is to let the intersection of the tangents to points $B$ and $C$ be $D$. Then $AD$ is the symmedian. Here is the proof I found on Wikipedia,

Let the reflection of $AD$ across the angle bisector of $\angle BAC$ meet $BC$ at $M'$. Then: $$\frac{BM'}{M'C} = \frac{AM'\frac{\sin\angle{BAM'}}{\sin\angle{ABM'}}}{AM'\frac{\sin\angle{CAM'}}{\sin\angle{ACM'}}} =\frac{\sin\angle{BAM'}}{\sin\angle{ACD}}\frac{\sin\angle{ABD}}{\sin\angle{CAM'}} =\frac{\sin\angle{CAD}}{\sin\angle{ACD}}\frac{\sin\angle{ABD}}{\sin\angle{BAD}} =\frac{CD}{AD}\frac{AD}{BD}=1$$

I don't understand the transition from the second to the third equation. It should be

$$\frac{BM'}{M'C} = \frac{AM'\frac{\sin\angle{BAM'}}{\sin\angle{ABM'}}}{AM'\frac{\sin\angle{CAM'}}{\sin\angle{ACM'}}}=\frac{\sin \angle BAM'\sin \angle ACM'}{\sin \angle ABM' \sin \angle CAM'}$$

but since the last fraction equals $\sin\angle{BAM'}\sin\angle{ACD}/\sin\angle{ABD}\sin\angle{CAM'}$ we should have $$\frac{\sin\angle{ACM'}}{\sin\angle{ABM'}}=\frac{\sin \angle ACD}{\sin \angle ABD}$$ But I don't have any idea how to prove it.

PNT
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    Note that $\angle DBC=\angle DCB=\angle A$. (Why?) Consequently, $\angle ABD=\angle A+\angle B=180^\circ-\angle C$ and so on ... – Blue May 29 '22 at 14:08

1 Answers1

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Let $E$ and $F$ be the projections of $D$ on $AB$ and $AC$ respectively. Since $DB=DC$ and $\widehat{DBC}=\widehat{BCD}=\widehat{A}$ we have that $DE/DF = \sin\widehat{EBD}/\sin\widehat{FDC}=\sin\widehat{C}/\sin\widehat{B}=c/b. $ This is enough to deduce that $D$ lies on the symmedian through $A$.

Jack D'Aurizio
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