2
$\begingroup$

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.

$\endgroup$
2
  • 2
    $\begingroup$ Explaining your difficulties may get you some answers. $\endgroup$
    – Narasimham
    Apr 11, 2016 at 8:00
  • $\begingroup$ 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? $\endgroup$
    – tlallstar6
    Apr 11, 2016 at 8:25

1 Answer 1

Reset to default
1
$\begingroup$

Hint: You have

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

Because they are vertically opposite angles.

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

$\endgroup$
8
  • $\begingroup$ $\triangle O_1 PS$ similar to $\triangle O_2 PR$? $\endgroup$
    – tlallstar6
    Apr 11, 2016 at 9:07
  • $\begingroup$ @tlallstar6 Yes, and you would have more information about the angles. Can you relate them? $\endgroup$
    – lEm
    Apr 11, 2016 at 9:08
  • $\begingroup$ $\angle O_1 SP$ = $\angle O_2 RP$ $\angle SO_1 P$ = $\angle RO_2 P$ $\angle SPR$ = $\angle O_1 PO_2$ $\triangle SPR$ 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. $\endgroup$
    – tlallstar6
    Apr 11, 2016 at 9:15
  • $\begingroup$ @tlallstar6 Actually you don't need to prove for similar triangles. You just need to show that some angles are equal in order to prove that the points are concyclic. (Think about what type of triangles they are) $\endgroup$
    – lEm
    Apr 11, 2016 at 9:18
  • $\begingroup$ Isosceles triangles as each of the two triangle has two equal sides (radius of the circle it is in) $\endgroup$
    – tlallstar6
    Apr 11, 2016 at 9:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .