# Help with Central Angle and Inscribed Angles

#### romsek

MHF Helper
Draw the bisector of the $50^\circ$ angle to the center.

This bisector is also a radius and thus forms two isosceles triangles that are each $(25^\circ, 25^\circ, 130^\circ)$ (make sure this is clear)

Thus the angular measure of the circumference that is not subtended by $x$ is $2 \times 130^\circ = 260^\circ$

Clearly $x + 260^\circ = 360^\circ$ and thus $x=100^\circ$

You will need to prove that the bisector does in fact pass through the center of the circle. Use symmetry arguments.

• 1 person

#### skeeter

MHF Helper
• 1 person

#### johng

Unfortunately, the "proof" provided by the link is fallacious. That "proof" requires that x + y is the measure of the inscribed angle PAQ. Tain't so. #### GLaw

Reflect Q in OA to the point Q′ on the other side of the circumference. Then $\angle\mathrm{PAQ^{\prime}}=x+y$.

#### metalworker

theorem, the angle which an arc of a circle subtends at the centre is double that which it subtends at the remaining part of the circumference,