Question

In the adjoining figure, O is the centre of the circle whose diameter is MN. Measures of some central angles are given in the figure. Hence find the following:
a.$m\angle AOB\,$ and $m\angle COD$
b.Show that arc $AB \cong $arc CD
c. Show that chord AB $ \cong $ chord CD.

Answer 1

Hint: In this question we have been given the angles in the circle. To solve this question we will use the concept which says that arcs corresponding to the chord of a circle are equal. We know that MN is the diameter of the circle, so we will also use the theorem that the sum of all the angles in a linear pair is of ${180^ \circ }$ .

Complete answer:
Let us solve the first in which we have to find the value of $m\angle AOB\,$ and $m\angle COD$
We can write that
 $\angle MOA + \angle AOB + \angle BON = 180$ , since all the angles are in the straight line.
By putting the values in the equation we have:
$ \Rightarrow 100 + \angle AOB + 35 = 180$
We will now simplify the equation:
$ \Rightarrow \angle AOB = 180 - (100 + 35)$
$ \Rightarrow \angle AOB = 180 - 135$
It gives us value:
$\angle AOB = 45$
Therefore we can say that $\angle COD = {45^ \circ }$ׄ.
In the second term, we can see that
\[ \Rightarrow \angle AOB = \angle COD = 45\]
Here the equal arcs subtended by equal angles at the centre are equal.
Hence we can say that $AB \cong $arc CD
Again we can apply the theorem which says that corresponding chords of congruent arcs are always congruent. Since $AB \cong $arc CD
Therefore we can say that chord AB$ \cong $ chord CD.

Note:
We should note the theorem which says that if two chords are equal in measure then their corresponding minor arcs are also equal in measure. We should know that the converse of this theorem is also true i.e. In a circle if two minor arcs are equal in measure then their corresponding chords are also equal in measure.