RD Chapter 16 Circles Ex 16.5 Solutions
Question - 21 : - Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.
Answer - 21 : -
Here, ABCD is a rhombus; we have to prove the four circles described on the four sides of any rhombus ABCD pass through the point of intersection of its diagonals AC and BD.
Let the diagonals AC and BD intersect at O.
We know that the diagonals of a rhombus intersect at right angle.
Therefore,
Now, means that circle described on AB as diameter passes through O.
Similarly the remaining three circles with BC, CD and AD as their diameter will also pass through O.
Hence, all the circles with described on the four sides of any rhombus ABCD pass through the point of intersection of its diagonals AC and BD.
Question - 22 : - If the two sides of a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are equal.
Answer - 22 : - 
To prove: AC = BD
Proof: We know that equal chords subtend equal at the centre of circle and the angle subtended by a chord at the centre is twice the angle subtended by it at remaining part of the circle.
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that:
(i) AD || BC
(ii) EB = EC.
Answer - 23 : -
(i) If ABCD is a cyclic quadrilateral in which AB and CD when produced meet in E such that EA = ED, then we have to prove the following, AD || BC
(ii) EB = EC
(i) It is given that EA = ED, so
Now
Question - 24 : - Circles are described on the sides of a triangle as diameters. Prove that the circles on any two sides intersect each other on the third side (or third side produced).
Answer - 24 : - 
Question - 25 : - Prove that the angle in a segment shorter than a semicircle is greater than a right angle.
Answer - 25 : - 
Prove that the angle in a segment shorter than a semicircle is greater than a right angle.
Question - 26 : - Prove that the angle in a segment greater than a semi-circle is less than a right angle.
Answer - 26 : - 
Question - 27 : - ABCD is a cyclic trapezium with AD || BC. If ∠B = 70°, determine other three angles of the trapezium.
Answer - 27 : -
If in cyclic quadrilateral 
, then we have to find the other three angles. Since, AD is parallel to BC, So,
(Alternate interior angles)
Now, since ABCD is cyclic quadrilateral, so
And,
Hence, ∠A=110°, ∠C=70° and ∠D=110°.
Question - 28 : - Prove that the line segment joining the mid-point of the hypotenuse of a right triangle to its opposite vertex is half the hypotenuse.
Answer - 28 : -
We have to prove that 
Let 
be a right angle at B and P be midpoint of AC Draw a circle with center at P and AC diameter
Since 
therefore circle passing through B So 
Hence 
Proved.