# Prove that points lie on the same circle

Let Circles $C_1$ (center $O_1$) and $C_2$ (center $O_2$) intersect at points $P$ and $Q$. $O_2$ is outside of $C_1$, $O_1$ is outside of $C_2$. Let $O_1P$ intersect $C_2$ at $R$ and $O_2P$ intersect $C_1$ at $S$. Prove that the points $R$, $S$, $O_1$, $O_2$ lie on the same circle.

Proving that the quadrilateral $O_1SRO_2$ can be inscribed in a circle seems like the next logical step, but I am having difficulties.

• Explaining your difficulties may get you some answers. Apr 11, 2016 at 8:00
• Apologies as I'm fairly new to this website. I have my vertical angles and I believe my next step is to move on to proving some similar triangles? Apr 11, 2016 at 8:25

Hint: You have

$$\angle O_1 PS =\angle O_2 PR$$

Because they are vertically opposite angles.

Then consider what type of triangles are $\triangle O_1 PS$

• $\triangle O_1 PS$ similar to $\triangle O_2 PR$? Apr 11, 2016 at 9:07
• $\angle O_1 SP$ = $\angle O_2 RP$ $\angle SO_1 P$ = $\angle RO_2 P$ $\angle SPR$ = $\angle O_1 PO_2$ $\triangle SPR$ similar to $\triangle O_1 PO_2$ $\angle PO_1 O_2$ = $\angle PSR$ $\angle PO_2 O_1$ = $\angle PRS$ Are these similar triangles due to SAS? I'm very poor at similar triangles. Apr 11, 2016 at 9:15