Question

Proved that the angle subtended on a semicircle is a right angle.

Answer 1

Hint: We start solving the problem by drawing the semicircle and connecting the ends to the point on the circumference of the circle. We join the points C and centre D to divide them into circles. We use the fact that angles opposite to equal sides are equal to get equal angles. We make calculations between the angles to get the required result.

Complete step-by-step answer:
Given that we need to prove the fact that angle subtended on a semicircle is the right angle.
Let us draw a semi-circle with endpoints A and B with centre D. Take a point ‘C’ on the circumference of the semicircle. Join ends of the semicircle with the point C to get the chords AC and BC. Now join the point ‘C’ and centre ‘D’.

We know that $ \angle ADC+\angle BDC={{180}^{o}} $ , as the angle of a straight line is $ {{180}^{o}} $ .
From triangle ADC, we see that the length of sides of DC and AD are equal as they both are the radius of the semicircle.
So, we get DC = AD. We know that angles opposite to equal sides are also equal.
So, we get $ \angle DAC=\angle ACD $ ---(1).
We know that the sum of all angles in triangle ADC is $ {{180}^{o}} $ .
We have got $ \angle ADC+\angle ACD+\angle CAD={{180}^{o}} $ .
From equation (1), we have $ \angle DAC=\angle ACD $ .
So, we have got $ \angle ADC+\angle ACD+\angle ACD={{180}^{o}} $ .
We have got $ \angle ADC+2\angle ACD={{180}^{o}} $ ---(2).
From triangle BDC, we see that the length of sides of DC and BD are equal as they both are the radius of the semicircle.
So, we get DC = BD. We know that angles opposite to equal sides are also equal.
So, we get $ \angle DBC=\angle BCD $ ---(3).
We know that the sum of all angles in triangle BDC is $ {{180}^{o}} $ .
We have got $ \angle BDC+\angle BCD+\angle CBD={{180}^{o}} $ .
From equation (3), we have $ \angle DBC=\angle BCD $ .
So, we have got $ \angle BDC+\angle BCD+\angle BCD={{180}^{o}} $ .
We have got $ \angle BDC+2\angle BCD={{180}^{o}} $ ---(4).
Let us add equations (2) and (4).
We have got $ \angle ADC+2\angle ACD+\angle BDC+2\angle BCD={{180}^{o}}+{{180}^{o}} $ .
We have got $ \angle ADC+\angle BDC+2\angle ACD+2\angle BCD={{360}^{o}} $ .
We know that $ \angle ADC+\angle BDC={{180}^{o}} $ and $ \angle ACD+\angle BCD=\angle ACB $ .
We have got $ {{180}^{o}}+2\angle ACB={{360}^{o}} $ .
We have got $ 2\angle ACB={{360}^{o}}-{{180}^{o}} $ .
We have got $ 2\angle ACB={{180}^{o}} $ .
We have got $ \angle ACB=\dfrac{{{180}^{o}}}{2} $ .
We have got $ \angle ACB={{90}^{o}} $ .
∴ The angle at point C subtended by AC and BC is $ {{90}^{o}} $ , which is a right angle.
We have proved that the angle subtended on a semicircle is the right angle.

Note: We should know that angles opposite to the equal sides are equal and vice-versa to solve many proofs. We should directly take triangles ADC and BDC as right angles triangles which are right angled at D. We should not assign values for angles at A, B and D to prove angle C.