I came across a question in my HW book:
Prove that an angle inscribed in a semicircle is a right angle.
My proof was relatively simple:
Proof: As the measure of an inscribed angle is equal to half the measure of its intercepted arc, the inscribed angle is half the measure of its intercepted arc, that is a straight line. As the arc's measure is $180^\circ$, the inscribed angle's measure is $180^\circ\cdot\frac{1}{2} = 90^\circ$.$\blacksquare$
When I checked the solution on the internet there were a whole bunch of other more complicated proofs. Is mine valid? If it is invalid, could someone tell me why?
Thank you,
Paul
Since $E$ is the midpoint of $E'E''$ and $\angle A=90^\circ$, $\triangle AEE'$ is isosceles at $E$. It follows that $\angle AEE''=2 \angle EAE'$, equivalently $E$ trisects the small arc $AC$.
Best Answer
Hint
Draw the radius from the center of the circle to the point that you think it has an angle of $90$ degrees and write down the angles:
$2(x+y)=180$