It is clearly seen that the angle at C is twice the angle at D. Here we have the fundamental theorem with regard to angles in circles -

" the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the circle "

This can be proved as follows :

Given a circle (C,r) in which arc AB subtends angle (ACB) at the centre and angle (ADB) at any point D on the circle.

To Prove:

(i) angle (ACB) = 2×angle (ADB), when arc AB is a minor arc or a semicircle.

(ii) Reflex angle (ACB)=2× angle (ADB), when arc AB is a major arc

Construction: Join AB and DC. Produce DC to a point X outside the circle.

Clearly, there are 3 cases.

Case I : arc AB is a minor arc [Fig (i)]

Case II : arc AB is a semicircle [Fig (ii)]

Case III : arc AB is a major arc [Fig (iii)]

We know that when one side of a triangle is produced then the exterior angle so formed is equal to the sum of the interior opposite angles.

Therefore,

angle (ACX) = angle (CAD) + angle (CDA) [consider triangle CAD]

angle (BCX) = angle (CBD) + angle (CDB) [consider triangle CBD]

But,

angle (CAD) = angle (CDA) [since CD=CA=r ] and ,

angle (CBD) = angle (CDB) [since CD=CB=r ]

Therefore,

angle (ACX)=2× angle (CDA) and

angle (BCX)=2×angle (CDB)

In Fig (i) and Fig (ii) ,

angle (ACX) + angle (BCX) = 2× angle (CDA) + 2× angle (CDB)⇒

angle (ACB) = 2[angle (CDA) + angle (CDB)] ⇒

angle (ACB) = 2×angle (ADB)

in Fig (iii),

angle (ACX) + angle (BCX) = 2× angle (CDA) + 2× angle (CDB)⇒

reflex angle (ACB) = 2[angle (CDA) + angle (CDB)] ⇒

reflex angle (ACB) = 2× angle (ADB)

Hence proved.

This theorem has 2 main implications:

(i) The angle in a semicircle is a right angle .

i.e. in fig (2), where arc AB is a semicircle, or in other words ACB is a diameter,

angle (ADB)=12× angle (ACB)=12×(180 degrees) = 90 degrees = a right angle

This holds true for any position of point C on the semicircle.

(ii) Angles in the same segment of a circle are equal.

This means that if an arc subtends 2 angles, at 2 different points on the circle, these angles will be equal.

We can prove this, by proving that each of the 2 angles is equal to half the angle subtended at the centre.