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If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Solution:
Concepts used to solve the question:

The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.

Diameter is the longest chord.
Let BD be the diameter of the circle, which is also a chord. Then, ∠BOD = 180°
Since AC and BD are diameters of the circle
⇒ AC = BD
We know that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
∴ ∠BAD = 1/2 × ∠BOD = 90°
Similarly, ∠BCD = 90°
Now, considering AC as the diameter of the circle, we get ∠AOC = 180°
We know that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
∠ABC = 1/2 × ∠AOC = 90°
Similarly, ∠ADC = 90°
Let us now consider the triangles ∆ABC and ∆BAD,
∠ABC = ∠BAD [Each equal to 90°]
AB = BA [Common side of triangle ABC and BAD]
AC = BD [Diameter of the circle]
∴ ∆ABC ≅ ∆BAD [By RHS congruence criteria]
⇒ BC = AD [By Corresponding parts of Congruent triangles theorem]
Similarly, AB = DC
As you can see, all the angles at the corners are 90º and the opposite sides are equal. We can say that the shape joining the vertices ABCD is a rectangle.
☛ Check: NCERT Solutions for Class 9 Maths Chapter 10
Video Solution:
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Maths NCERT Solutions Class 9 Chapter 10 Exercise 10.5 Question 7
Summary:
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, then it is a rectangle.
☛ Related Questions:
 If the nonparallel sides of a trapezium are equal, prove that it is cyclic.
 Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D, P and Q respectively (see Fig. 10.40). Prove that ∠ACP = ∠QCD.
 If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
 ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD
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