# Geometry question from Triangles

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 $$\frac{SD}{S'D}=\frac{BD}{DC}.$$

• Don't use the tag proof-verification if you are not asking for one. – user10354138 May 23 '19 at 18:45

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.

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

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

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

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