본문으로 바로 이동 / Skip to content
MathIsland
한EN

PROOF 52

Prove.

The angle in a semicircle is a right angle

On the Euclidean plane.

A way to read this proof

First separate what is assumed from what is to be proved. At each step, ask which definition or earlier result permits the move. If it is unclear, return to that premise rather than skipping to the conclusion.

Definition D1

In the circle with center O and radius r, let AB be a diameter and C a point on the circle distinct from A and B. The angle ∠ACB is called the inscribed angle on the diameter.

Proof.

(1)O lies in the interior of segment AB, and AO = OB = rconvention
(2)C is not on line AB, so A, B, C form a triangle.convention

∎

Theorem T1

If AB is a diameter and C is a point on the circle, then ∠ACB = 90°

Proof.

(1)OA = OB = OC = rD1, A1
(2)In triangle OAC, OA = OC, so ∠OAC = ∠OCA. Call this value α.(1), A2
(3)In triangle OBC, OB = OC, so ∠OBC = ∠OCB. Call this value β.(1), A2
(4)O lies in the interior of segment AB, so ray CO is interior to ∠ACB.D1, A6
(5)∠ACB = ∠ACO + ∠OCB = α + β(4), A5, (2), (3)
(6)Ray AO is the same as ray AB and ray BO the same as ray BA, so ∠CAB = α, ∠CBA = βD1, (2), (3)
(7)In triangle ABC, α + β + (α + β) = 180°(5), (6), A3
(8)2(α + β) = 180°(7), A8
(9)α + β = 90°(8), A8, E
(10)∠ACB = 90°(5), (9), E

∎

Theorem T2

The same conclusion follows from the inscribed angle theorem of Proof 51.

Proof.

(1)∠ACB is the inscribed angle on the arc AB not containing C, and that arc is a semicircle.D1
(2)O is between A and B, so the central angle of that arc is ∠AOB = 180°D1, A4
(3)∠ACB = ½ × 180°(1), (2), A7
(4)∠ACB = 90°(3), A8

∎

Remark

Can A7 be removed from this proof?

The two theorems reach the same conclusion. Separate what each one rests on.