Question : Prove that if you draw a triangle inside a semicircle, the angle opposite the diameter is 90°.
Proof :
Label the diameter endpoints A and B, the top point C and the middle of the circle M.
Label the acute angles at A and B Alpha and Beta.
Draw a radius 'r' from the (right) angle point C to the middle M.
Angle MAC = ACM = Alpha because the left subtriangle is iscosceles because the opposite sides AM and CM are both radii.
Angle MBC = BCM = Beta because the right subtriangle is iscosceles because the opposite sides BM and CM are both radii.
Add up the angles at A, B and C.
This gives 2 * Alpha + 2 * Beta, which sum to 180° because ABC is a triangle.
Halving this gives Alpha + Beta (= the angle ACB) = 90°
Q.E.D :-)
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