# Circumscribed circle

$$\triangle ABC: \angle CAB = 45^o, BC = 9$$. I have to find the diameter of the circumscribed circle. I thought about an hour, but I can't think up anything. I would be very grateful if you can help me!

$$\triangle OBC$$ is right angled, and $$OB=OC$$. You'll be able to find the radius.

• Thank you! I got it. I appreciate your help. – Nikol Dimitrova Apr 15 at 17:30
• You're welcome$\quad$ – HAMIDINE SOUMARE Apr 15 at 17:31

Hint: Use that $$\sin(\alpha)=\frac{a}{2R}$$

• I am in 8th grade and I don't know sin. – Nikol Dimitrova Apr 15 at 17:22

Let the perpendicular on $$BC$$ drawn from $$O$$ meet $$BC$$ at the point $$D.$$ Then observe that $$\Delta OBD \cong \Delta OCD.$$ So $$BD=CD = \frac 9 2.$$ Also observe that $$\angle BOC = 2 \times \angle BAC = 90^{\circ}.$$ So $$\angle OBC = \angle OCB = 45^{\circ}.$$ Now let $$OC=r.$$ Then we have $$r=\frac {\frac 9 2} {\cos 45^{\circ}} = \frac {9} {\sqrt 2}.$$

So the diameter of the circle is $$2r = 9\sqrt 2.$$

Another approach $$:$$ Since $$\angle BOC = 90^{\circ}$$ so $$\Delta OBC$$ is a right angled triangle. Let $$r$$ be the radius of the circle. Now observe that $$OB=OC = r.$$ Then by Pythagoras theorem we have $$2r^2 =81 \implies r = \frac {9} {\sqrt 2}.$$ So the diameter of the circle is $$2r = 9 \sqrt 2.$$