CAT Cyclic Quadrilaterals - Practice Questions & MCQ

Theorem 10 : If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e. they are concyclic).

You can see the truth of this result as follows:

AB is a line segment and C, D are two points lying on the same side of AB such that ∠ ACB = ∠ ADB.

We have to show that the points A, B, C and D lie on a circle. 

Let us draw a circle through the points A, C and B.

If D lies on the circle passing through A, B and C then clearly the result follows.

If possible, suppose D does not lie on this circle.

Then, this circle will intersect AD or AD produced in D'.

Now,                ∠ ACB = ∠ ADB             (given)

Now,               ∠ ACB = ∠ AD'B             ([angles in the same segment)

Therefore,      ∠ ADB = ∠ AD'B

But, an exterior angle of a triangle can never be equal to its interior opposite angle.

So, ∠ ADB = ∠ AD'B is true unless D' coincide with D.

Thus, D lies on the circle passing through A, B and C.

Hence, the points A, B, C, D are concyclic.

A quadrilateral ABCD is called cyclic if all the four vertices of it lie on a circle.

Theorem 11 : The sum of either pair of opposite angles of a cyclic quadrilateral is 180º.

Let's prove this

We have given a cyclic quadrilateral ABCD

We have to prove ∠A + ∠C = 180° and ∠B + ∠D = 180°

Join AC and BD

Consider side AB of quadrilateral ABCD as the chord of the circle. We have,

                     ∠ACB = ∠ADB                ...(1)  (angle in the same segment are equal)

 and              ∠BAC = ∠BDC                ...(2)  (angle in the same segment are equal)

Add (1) and (2)

We get,             ∠ACB + ∠BAC = ∠ADB + ∠BDC

This Gives,        ∠ACB + ∠BAC = ∠ADC

Add ∠ABC on both side, we get

            ∠ACB + ∠BAC + ∠ABC = ∠ADC + ∠ABC

Since,  ∠ACB + ∠BAC + ∠ABC = 180°        (the sum of the angle of ∆ABC is 180°) 

So,                    ∠ADC + ∠ABC = 180° 

                                   ∠B + ∠D = 180° 

But,           ∠A + ∠B + ∠C + ∠D = 360°

Therefore,           ∠A + ∠C = 360° - (∠B + ∠D) 

                           ∠A + ∠C = 360° - 180° = 180°

Hence,  ∠A + ∠C = 180° and ∠B + ∠D = 180°.

The converse of the above theorem is

Theorem 12 : If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic.

Theorem 13 : The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Let's prove this,

Given: ABCD is a cyclic quadrilateral.

Construction: Extend BC to X.

To Prove: ∠DCX = ∠BAD.

∠BAD + ∠BCD = 180°                ...(1)

(The opposite angles of a cyclic quadrilateral are supplementary).

∠BCD + ∠DCX = 180°               ...(2)

(Angle of a straight line)

From Eqs. (1) and (2), we get

∠BAD + ∠BCD = ∠BCD + ∠DCX

⇒ ∠DCX = ∠BAD.