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Question
Prove by vector method, that the angle subtended on semicircle is a right angle.
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Solution 1
Let seg AB be a diameter of a circle with centre C and P be any point on the circle other than A and B.
Then ∠APB is an angle subtended on a semicircle.
Let `bar"AC" = bar"CB" = bar"a"` and `bar"CP" = bar"r"`
Then `|bar"a"| = |bar"r"|` ....(1)
`bar"AP" = bar"AC" + bar"CP"`
= `bar"a" + bar"r"`
= `bar"r" + bar"a"`
`bar"BP" = bar"BC" + bar"CP"`
= `- bar"CB" + bar"CP"`
= `- bar"a" + bar"r"`
∴ `bar"AP".bar"BP" = (bar"r" + bar"a").(bar"r" - bar"a")`
= `bar"r".bar"r" - bar"r".bar"a" + bar"a".bar"r" - bar"a".bar"a"`
= `|bar"r"|^2 - |bar"a"|^2`
= 0 ....`(∵ bar"r".bar"a" = bar"a".bar"r")`
∴ `bar"AP" ⊥ bar"BP"`
∴ ∠APB is a right angle.
Hence, the angle subtended on a semicircle is the right angle.
Solution 2
Consider the circle with the centre at O and AB is the diameter.
Let `bar(OA) = bar a, bar(OB) = bar b, bar(OC) = bar c`
∴ `|bar a| =|bar b| = |bar c| = r` ...(1)
and `bar a = -bar b` ...(2)
Consider:
`bar (AC) * bar (BC) = (bar c - bar a) * (bar c - bar b)`
= `(bar c - bar a) * (bar c + bar a)` ...[From (2)]
= `|bar c|^2 - |bar a|^2`
= r2 − r2 ...[From (1)]
= 0
∴ `bar(AC) * bar(BC) = 0`
∴ `bar(AC)` is perpendicular to `bar(BC)`
∴ ∠ACB = 90°
∴ Angle subtended on semi-circle is a right angle.
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