Thales' Theorem/Converse

From ProofWiki
Jump to navigation Jump to search

Theorem

Thales' theorem converse.png

Let $O$ be a circle.

Let $AOB$ be a diameter of $O$.

Then $\angle APB$ is a right angle if and only if $P$ lies on the circle $O$.


Proof

Necessary Condition

Draw $OC \parallel PB$.

By Parallelism implies Equal Corresponding Angles:

$\angle ACO = \angle APB$ and both are right angles.

$\triangle ACO$ and $\triangle APB$ share $\angle OAC$.

$\triangle ACO$ and $\triangle APB$ are both right triangles.

\(\ds \triangle ACO\) \(\sim\) \(\ds \triangle APB\) Triangles with Two Equal Angles are Similar
\(\ds AO : AB\) \(=\) \(\ds AC : AP\) Definition of Similar Triangles
\(\ds AO\) \(=\) \(\ds \frac 1 2 AB\) Definition of Radius of Circle
\(\ds AC\) \(=\) \(\ds \frac 1 2 AP\) Definition of Similar Triangles
\(\ds AC\) \(=\) \(\ds CP\)

Now draw $OP$.

$OC$ is shared.

\(\ds \angle ACO\) \(=\) \(\ds \angle PCO\) both are right angles
\(\ds \triangle ACO\) \(\cong\) \(\ds \triangle PCO\) Triangle Side-Angle-Side Congruence
\(\ds OP\) \(=\) \(\ds OA\)

But $OA$ is a radius of the circle.

Therefore $OP$ is also a radius.

So $P$ must lie on the circle.

$\Box$

Sufficient Condition

Thales-Theorem.png

Given:

$AO = CO = BO$

By definition of isosceles triangles:

$\triangle AOC$ and $\triangle BOC$ are isosceles triangles
\(\ds \angle OAC + \angle OCA + \angle OCB + \angle OBC\) \(=\) \(\ds 180^{\circ}\) Sum of Angles of Triangle equals Two Right Angles
\(\ds \angle OAC\) \(=\) \(\ds \angle OCB\) Isosceles Triangle has Two Equal Angles
\(\ds \angle OCB\) \(=\) \(\ds \angle OBC\) Isosceles Triangle has Two Equal Angles
\(\ds 2 (\angle OCA + \angle OCB)\) \(=\) \(\ds 180^{\circ}\) Substitution
\(\ds \angle OCA + \angle OCB\) \(=\) \(\ds 90^{\circ}\)

The result follows.

$\blacksquare$


Also see

The Sufficient Condition is Thales' Theorem.

Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Thales%27_Theorem/Converse&oldid=659127"