The right-angled triangle inscribed in a circle

In the general case, when a chord of a circle intercepts two sides of an inscribed triangle, the opposite angle formed is equal to half the arc formed by the chord.

Un triangle rectangle inscrit dans un cercle with un côté comme corde du cercle

$$ \alpha = \frac{\overset{\frown}{BC}}{2} $$

As in the case where the chord is the diameter of the circle, the arc $ \overset{\frown}{BC} \enspace = \pi $, such as the following figure:

Un triangle rectangle inscrit dans un cercle

Thus $ \alpha = \frac{\pi}{2}$, and this triangle will be right-angled.

In a circle, if an inscribed triangle has the diameter of this circle as its longest side, then this triangle will be right-angled.

Demonstration

To do this, let us consider an orthonomous coordinate system $ (0, \vec{x}, \vec{y}) $.

Démonstration du triangle rectangle inscrit dans un cercle

We called the sides of the triangle $ a, b, c $, and we projected a height $ h $ along the length $ c $, with as a point of intersection $ x_R $ .

Likewise, we traced the radius $ R $ of the circle pointing towards the vertex between $ a $ and $ b $ of the triangle.

We will seek to demonstrate that the triangle $ \{a, b, c \} $ is right-angled between both sides $ a $ and $ b $.

We know from the Pythagorean Theorem reciprocal that:

$$ a^2 + b^2 = c^2 \Longrightarrow a \perp b \qquad \bigl(Pythagore \enspace(reciprocal) \bigr) $$

C'est ce que nous allons démontrer.

We apply the Pythagorean Theorem on three right-angled triangles:

- that formed by $ \{h, (R - x_r), a \} $

- that formed by $ \{h, (R + x_r), b\} $

- that formed by $ \{h, x_R, R \} $

We then have the following equalities:

$$ h^2 + (R - x_R)^2 = a^2 \qquad (1) $$

$$ h^2 + (R + x_R)^2 = b^2 \qquad (2) $$

$$ h^2 + x_R^2 = R^2 \Longleftrightarrow h^2 = R^2 - x_R^2\qquad (3) $$

Let us inject the expression of $ h^2 $ of the expression $ (3) $ into $ (1) $ and $ (2) $, we then obtain a new pair of equality:

$$ \Biggl \{ \begin{gather*} R^2 - x_R^2 + (R - x_R)^2 = a^2 \qquad (4) \\ R^2 - x_R^2 + (R + x_R)^2 = b^2 \qquad (5) \end{gather*} $$

Let us now calculate $ a^2 + b^2 $ by performing $ (4) + (5) $:

$$ a^2 + b^2 = R^2 - x_R^2 + (R - x_R)^2 + R^2 - x_R^2 + (R + x_R)^2 $$

$$ a^2 + b^2 = R^2 - x_R^2 + R^2 - 2R.x_R + x_R^2 + R^2 - x_R^2 + R^2 + 2R.x_R + x_R^2 $$

Many terms cancel, and we obtain:

$$ a^2 + b^2 = 4R^2 $$

$$ a^2 + b^2 = (2R)^2 $$

Moreover, we know that:

$$ c = 2R $$

Because it is our diameter by hypothesis.

And finally,

$$ a^2 + b^2 = c^2 $$

With the Pythagorean Theorem reciprocal, we showed that the triangle $ \{a, b, c \} $ is right-angled between the sides $ a $ and $ b $.

Thus,

In a circle, if an inscribed triangle has the diameter of this circle as its longest side, then this triangle will be right-angled.