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Suppose $AD$ bisects angle $A$ of triangle $ABC$ and meets $BC$ at $D$, and let $S$ and $S'$ be the circumcenters of triangles $ABD$ and $ACD$ respectively. Show that $$\frac{SD}{S'D}=\frac{BD}{DC}.$$

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  • $\begingroup$ Don't use the tag proof-verification if you are not asking for one. $\endgroup$ – user10354138 May 23 '19 at 18:45
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Using the law of sines,

$SD=\frac{BD}{2\sin\angle BAD}$

$S'D=\frac{CD}{2\sin\angle DAC}$

and since $\angle BAD=\angle DAC$ because $AD$ is an angle bisector, we get the result you wanted.

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$\alpha = \beta$ because all green-shaded angles are equal.

enter image description here

Then, $\triangle ABC \sim \triangle DSS’$

Therefore, $\dfrac {SD}{S’D} = \dfrac {AB}{AC}$

But, by bisector theorem, $\dfrac {AB}{AC} = \dfrac {BD}{DC}$.

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