An inscribed angle on a diameter

Theorem: In a circle, an inscribed angle standing on a diameter is a right angle ($\displaystyle 90^{\circ}$).

Given: $\displaystyle AB$ is a diameter of the circle with centre $\displaystyle O$, and $\displaystyle C$ is on the circle.
To prove: $\displaystyle {\angle ACB}={90^{\circ}}$.

The proof below shows that the central angle $\displaystyle \angle AOB$ is a straight angle, and uses the theorem of the last lesson – an inscribed angle is half the central angle:

Theorem: In a circle, an inscribed angle of $\displaystyle 90^{\circ}$ stands on a diameter.

• An inscribed angle of $\displaystyle 90^{\circ}$ stands on an arc of $\displaystyle 180^{\circ}$ – half the circle.
• The chord of half a circle passes through the centre, so it is a diameter.

The circle $\displaystyle {(x-2)^{2}+(y-5)^{2}}={25}$ has its centre at $\displaystyle M(2,5)$ and radius $\displaystyle 5$.
The points are $\displaystyle A(6,2)$, $\displaystyle B(-2,8)$ and $\displaystyle C(5,9)$.

• $\displaystyle A$ is on the circle: $\displaystyle {(6-2)^{2}+(2-5)^{2}}={16+9}={25}$. So are $\displaystyle B$ and $\displaystyle C$.
• $\displaystyle AB$ is a diameter: the midpoint of $\displaystyle AB$ is $\displaystyle {\left(\frac{6+(-2)}{2},\frac{2+8}{2}\right)}={(2,5)}$ – the centre.
• The slopes: $\displaystyle {m_{AC}}={\frac{9-2}{5-6}}={-7}$ and $\displaystyle {m_{BC}}={\frac{9-8}{5-(-2)}}={\frac{1}{7}}$.
• Their product is $\displaystyle {-7\cdot\frac{1}{7}}={-1}$, so $\displaystyle AC\perp BC$: the angle $\displaystyle \angle ACB$ is right, as the theorem says.

In the circle $\displaystyle {(x-2)^{2}+(y-5)^{2}}={25}$ of the example, $\displaystyle AB$ is the diameter: $\displaystyle A(6,2)$, $\displaystyle B(-2,8)$.
The point $\displaystyle D$ is $\displaystyle (5,1)$.

How big is $\displaystyle \angle ADB$? Give your reason.

From the converse: in a right-angled triangle, the hypotenuse is a diameter of the circle circumscribed about the triangle – the circle through its three vertices.

• The centre of the circle is the midpoint of the hypotenuse, and the radius is half the hypotenuse.
• This is grade 10’s property too: the median to the hypotenuse is half the hypotenuse.

The legs of a right-angled triangle are $\displaystyle 6$ cm and $\displaystyle 8$ cm.
The triangle is inscribed in a circle.

$\displaystyle AB$ is a diameter, and $\displaystyle C$ is on the circle.
$\displaystyle {\angle CAB}={35^{\circ}}$.
Find $\displaystyle \angle CBA$.

$\displaystyle AB$ and $\displaystyle EF$ are two diameters of the circle with centre $\displaystyle O$.
Prove: the quadrilateral $\displaystyle AEBF$ is a rectangle.