Question

In the given figure, if ABCD is a rhombus, diagnosis AC and BD intersect at O and D is a point lying on the circle with centre O, then the sum of the measures of $\angle$BAE and $\angle$EDC is equal to.

• A. ${180}^o$
• B. ${60}^o$
• C. ${90}^o$
• D. ${45}^o$

Answer 1

The correct option is D ${45}^o$

REF.Image.

Given :

ABCD is a rhombus

AC and BD interest are O

E any point on circle

with centre O.

To field :- Sum of measure

of $\angle BAE$ and $\angle EDC$

$So{l}^n$ :- Since ABCD is a rhombus

and diagonals of rhombus bisect at ${90}^{\circ }$

So, $\angle BOC={90}^{\circ }$

and similarly $\angle DOA={90}^{\circ }$

Now we have a shown that angle drawn on the centre

of the circle by the same are is qoulde the ...(1)

angle drawn on any other point on the circle

So, it from the are $BEC$ angles $\angle BAC=\frac{1}{2}\angle BOC.$

and $\angle BOC={90}^{\circ }$

So $\angle BAC=\frac{{90}^{\circ }}{2}={45}^{\circ }$

from the same theorem we have that $\angle BDC={90}^{\circ }...\left(ii\right)$

(drawn from same are BEC)

Now we have a theorem that angles drawn from same

are of a circle are equal

$\Rightarrow$ angle from are $BE\to$ means $\angle BAE=\angle BDE...\left(iii\right)$

(drawn from same are BE)

again similarly from are $EC\to \angle EDC=\angle EAC...\left(iv\right)$

(drawn from same are EC).

from (1) $\angle BAE=\angle BDE$

adding $\angle EDC$ both side

$\angle BAE+\angle EDC=\angle BDE+\angle EDC=\angle BDC={45}^{\circ }$