# Question #0583a

Mar 14, 2017

see explanation.

#### Explanation:

Given that $\angle A B C = {90}^{\circ}$,
$\implies A C$ is the diameter of the circumscribed circle of $\Delta A B C$.
As $B E$ is perpendicular to diameter $A C$,
$\implies B D = D E$

Intersecting chord theorem : when two chords intersect each other inside a circle, the products of their segment are equal.
$\implies A D \cdot D C = B D \cdot D E = B {D}^{2}$
$\implies 16 \cdot 9 = B {D}^{2} , \implies B D = 12$.

By Pythagorean Theorem:
${x}^{2} = A {D}^{2} + B {D}^{2} = {16}^{2} + {12}^{2} = 400$
$\implies x = \sqrt{400} = 20$

${y}^{2} = C {D}^{2} + B {D}^{2} = {9}^{2} + {12}^{2} = 225$
$\implies y = \sqrt{225} = 15$

Hence, $B D = 12 , x = 20 , y = 15$

Mar 15, 2017

Sol. 2. (see explanation).

#### Explanation:

Solution 2.

As kindly suggested by Ratnaker Mehta, the following results are very useful in such geometrical problems.

(1) $B {D}^{2} = A D \cdot D C$
(2) $A {B}^{2} = A D \cdot A C$
(3) $C {B}^{2} = C D \cdot C A$

$\implies B {D}^{2} = 16 \cdot 9 = 144 , \implies B D = \sqrt{144} = 12$
$\implies A {B}^{2} = 16 \cdot \left(16 + 9\right) = 400 , \implies A B = \sqrt{400} = 20$
$\implies C {B}^{2} = 9 \cdot \left(16 + 9\right) = 225 , \implies C B = \sqrt{225} = 15$

proof :

$\Delta A B C , \Delta A D B \mathmr{and} \Delta B D C$ are similar triangles.

(1) $\frac{A D}{D B} = \frac{B D}{D C} , \implies B {D}^{2} = A D \cdot D C$

(2) $\frac{A B}{A C} = \frac{A D}{A B} , \implies A {B}^{2} = A D \cdot A C$

(3) $\frac{B C}{C A} = \frac{D C}{C B} , \implies B {C}^{2} = C D \cdot C A$