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Every Point on the Bisector of an Angle Is Equidistant from the Sides of the Angle.

AB is a diame...

Question

$A B$ is a diameter of a circle. $P$ is a point on the semi-circle $A P B$ . $A H$ and $B K$ are perpendiculars from $A$ and $B$ respectively to the tangent at $P$ . Prove that $A H + B K = A B$ .

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Solution

Clearly, $∠ M P O = 90^{o}$

Since $B K ⊥ H M, A H ⊥ H M$ and $O P ⊥ H M$ . Therefore,
$A H ∥ O P ∥ B K$

Let $A H = x, B K = y$ and $O P = r$ . Further, let $B M = z$ .

In $△ M K B$ and $△ M H A$ , we have
$∠ M K B = ∠ M H A = 90^{o}$

and $∠ B M K = ∠ A M H$

So, by $A A$ criterion of similarity, we have

$△ M K B \sim △ M H A$

$\Rightarrow \frac{B K}{A H} = \frac{M B}{M A}$

$\Rightarrow \frac{y}{x} = \frac{z}{2 r + z} \Rightarrow 2 r y + y z = z x \Rightarrow z = \frac{2 r y}{x - y}$ . . . $(i)$

In $△ M K B$ and $△ M P O$ , we have

$∠ M K B = ∠ M P O = 90^{o}$

$∠ B M K = ∠ O M P$

So, by $A A$ criterion of similarly, we have

$△ M K B \sim △ M P O$

$\Rightarrow \frac{B K}{O P} = \frac{B M}{O M}$

$\Rightarrow \frac{y}{r} = \frac{z}{r + z}$

$\Rightarrow r y + y z = r z$

$\Rightarrow z = \frac{r y}{r - y}$ . . . $(i i)$

From $(i)$ and $(i i)$ , we get

$\frac{r y}{r - y} = \frac{2 r y}{x - y}$

$\Rightarrow 2 r - 2 y = x - y$

$\Rightarrow 2 r = x + y$

$\Rightarrow A B = A H + B K$

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AB is a diameter of a circle. P is a point on the semi-circle APB. AH and BK are perpendiculars from A and B respectively to the tangent at P. Prove that AH+BK=AB.

$A B$ is a diameter of a circle. $P$ is a point on the semi-circle $A P B$ . $A H$ and $B K$ are perpendiculars from $A$ and $B$ respectively to the tangent at $P$ . Prove that $A H + B K = A B$ .