# Proof that the angle of a triangle in a semicircle is 90°.

## © Stu Savory, 2006.

**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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