Question
Prove by vector method, that the angle subtended on semicircle is a right angle.
✓
Answer
Let seg $A B$ be a diameter of a circle with centre $C$ and $P$ be any point on the circle other than $A$ and $B$.
Then $\angle A P B$ is an angle subtended on a semicircle.
Let $\overline{ AC }=\overline{ CB }=\overline{ a }$ and $\overline{ CP }=\overline{ r }$
Then $|\overline{ a }|=|\overline{ r }|$$\ldots(1)$
$ \overline{ AP }=\overline{ AC }+\overline{ CP }$
$=\overline{ a }+\overline{ r }$
$=\overline{ r }+\overline{ a }$
$\overline{ BP }=\overline{ BC }+\overline{ CP }$
$=-\overline{ CB }+\overline{ CP }$
$=-\overline{ a }+\overline{ r }$
$\therefore \overline{ AP } \cdot \overline{ BP }=(\overline{ r }+\overline{ a }) \cdot(\overline{ r }-\overline{ a })$
$=\overline{ r } \cdot \overline{ r }-\overline{ r } \cdot \overline{ a }+\overline{ a } \cdot \overline{ r }-\overline{ a } \cdot \overline{ a }$
$=|\overline{ r }|^2-|\overline{ a }|^2$
$=0 \quad \ldots .(\because \overline{ r } \cdot \overline{ a }=\overline{ a } \cdot \overline{ r })$
$\therefore \overline{ AP } \perp \overline{ BP } $
$\therefore \angle APB$ is a right angle.
Hence, the angle subtended on a semicircle is the right angle.
Then $\angle A P B$ is an angle subtended on a semicircle.
Let $\overline{ AC }=\overline{ CB }=\overline{ a }$ and $\overline{ CP }=\overline{ r }$
Then $|\overline{ a }|=|\overline{ r }|$$\ldots(1)$
$ \overline{ AP }=\overline{ AC }+\overline{ CP }$
$=\overline{ a }+\overline{ r }$
$=\overline{ r }+\overline{ a }$
$\overline{ BP }=\overline{ BC }+\overline{ CP }$
$=-\overline{ CB }+\overline{ CP }$
$=-\overline{ a }+\overline{ r }$
$\therefore \overline{ AP } \cdot \overline{ BP }=(\overline{ r }+\overline{ a }) \cdot(\overline{ r }-\overline{ a })$
$=\overline{ r } \cdot \overline{ r }-\overline{ r } \cdot \overline{ a }+\overline{ a } \cdot \overline{ r }-\overline{ a } \cdot \overline{ a }$
$=|\overline{ r }|^2-|\overline{ a }|^2$
$=0 \quad \ldots .(\because \overline{ r } \cdot \overline{ a }=\overline{ a } \cdot \overline{ r })$
$\therefore \overline{ AP } \perp \overline{ BP } $
$\therefore \angle APB$ is a right angle.
Hence, the angle subtended on a semicircle is the right angle.
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