# Find $∠ADC$ given $A,D$ on circle with diameter $BC$ and $∠BAO = 50°$

If $$BC$$ is a diameter of the circle and $$∠BAO = 50°$$. Then find the value of $$∠ADC$$.

This stumped me a little, I think there is a rule that mentions something about the center of the circle and it's relation to angles on the circumfrence, but I'm not sure, could someone help me point out the laws that would help to solve this problem? (Do not give me the solution, since I want to approach this myself)

Thanks in advance! :)

• en.wikipedia.org/wiki/Inscribed_angle Feb 14, 2021 at 7:12
• But frankly, you don't really need this. The problem can be done just using the fact that the sum of angles in a triangle is $180^\circ$ and a few basic things about angles. Feb 14, 2021 at 7:26
• The question isn't well formulated. There's no restriction on the location of D and it could be anywhere in the cicle. Therefore, the are an infinite number of values of ADC that would meet the criteria. Feb 14, 2021 at 15:27

## Hints

Mark the congruent line segments in the figure (a circle has a constant radius) to observe that:

$$\measuredangle ABC = \measuredangle BAO$$

Then, try to use the fact that the measures of the angles in the same segment of a circle are equal (the measures of two angles at the circumference subtended by the same arc are equal):

$$\measuredangle ABC = \measuredangle ADC$$

• ABC = ADC? I don't think so. Feb 14, 2021 at 15:23
• @TheImpaler Why don't you think so? The measures of the angles are clearly equal. Feb 14, 2021 at 15:43
• Because AD is not a diameter. Feb 14, 2021 at 22:56

See $$\angle OAB = \angle OBA = 50^\circ$$ (why?)

$$\angle AOC = 100^\circ$$ (why?)

Now see angle subtended in the centre is double the angle subtended at any other point on the circle. Join AC. Now $$\angle AOC = 2\angle ADC$$

$$\angle ADC= 50^\circ$$

• You have a small mistake. It should be $\angle AOC = 2 \angle ADC$ instead of $2\angle AOC = \angle ADC$ Feb 14, 2021 at 8:24
• Oh I didn't realised that. Thanks for pointing the mistake
– user876009
Feb 14, 2021 at 8:25

Since $$AO=OB$$, $$\angle A =\angle B=50^\circ$$. Thus $$\angle O =100^\circ$$. Since $$\angle ADC= \frac1 2 \angle AOC$$,

$$\angle ADC= 50^\circ$$.

• AO = AB? There's no indication of this in the question. Feb 14, 2021 at 15:28
• @TheImpaler Happy? Feb 14, 2021 at 16:02