# Please solve q 80 ?

May 17, 2018

option (4) is acceptable.

#### Explanation:

Given that, $A B = A C = B D$ and $A C \bot B D$.

$\rightarrow A B = A C$

$\rightarrow \angle B = \angle C$

$\rightarrow 90 - a + 90 - d = d$

$\rightarrow a = 180 - 2 d$.....[1]

Also, $\rightarrow A B = B D$

$\rightarrow \angle A = \angle D$

$\rightarrow a + b = 90 - b$

$\rightarrow a = 90 - 2 b$....[2]

From [1] and [2], we have,

$\rightarrow 180 - 2 d = 90 - 2 b$

$\rightarrow d - b = 45$....[3]

Now, $\angle C + \angle D$

$= \angle B C A + \angle B D A = 90 - b + d = 90 + 45 = 135$

May 19, 2018

$\angle C + \angle D = {135}^{\circ}$

#### Explanation:

Let $\angle A B D = x$,
given $A B = B D , \implies \angle B D A = \frac{180 - x}{2}$
let $\angle C B D = y$,
given $A B = A C , \implies \angle A C B = \angle A B C = x + y$,
In $\Delta C B E , \implies \textcolor{red}{x + 2 y = {90}^{\circ}}$
$\angle C + \angle D = \angle A C B + \angle B D A = x + y + \frac{180 - x}{2}$
$= x + y + 90 - \frac{x}{2}$
$= \frac{x}{2} + y + 90$
$= \frac{x + 2 y}{2} + 90$
$= \frac{90}{2} + 90 = 45 + 90 = {135}^{\circ}$