Suppose we draw circle with the bottom side of the triangles in the picture as diameter. Find out whether the top corner of each triangle is inside the circle, on the circle or outside the circle.

● To see how right triangle makes a circle?

1. Draw a line of 5 cm.

2. Using Line with given length, make a slider from 0 to 180° with increment 5.

3. Draw an angle ‘a’ at one end and draw ‘90° - a’ clockwise at the other end to make a triangle.

4. Draw several such triangles.

5. Apply trace on to the top vertex and sides of a triangle to get a circle.

● To see angle in a semicircle is right.

1. To show ∠APB = 90° . Firstly join P to the centre of a circle O which splits the angle at P into two parts.

2. Δ AOP and Δ BOP are isosceles triangle.

Thus, ∠A = x° and ∠B = y°

3. In ΔAPB, the sum of angle is 180°

⇒ x + y + (x + y) = 180°

⇒ 2(x + y) = 180°

⇒ x + y = 90°

⇒ ∠APB = x° + y° = 90°

Thus, angle in a semicircle is right angle.

● To see for any point inside the circle, such an angle is larger than right angle.

1. To show ∠APB > 90° . Extend one of the lines to meet the circle. Join this point to the other end of the diameter.

2. APB is the exterior angle at P of triangle AQP.

∠APB = sum of interior angle at Q and B

3. The angle at Q is right angle. So, ∠APB is larger than 90°

● To see for any point outside the circle, such an angle is smaller than right angle.

1. To show ∠APB < 90° . APB is the interior angle at P of triangle PQB. The right angle AQB is an exterior angle.

2. So, ∠APB is smaller than 90°

Now, in our given question we can see that,

As seen above, point on a circle, such an angle is a right angle.

Point outside the circle , such an angle is larger than 90°

Point inside the circle, such an angle is smaller than 90°

Thus, as ∠ACB = 110° > 90° . Therefore point C lies inside the circle.

As, ∠ADB = 90° . Therefore point D lies on the circle.

As, ∠AEB = 70° . Therefore point E lies outside the circle.