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# Prove that the sum of opposite sides of a cyclic quadrilateral is $\pi$

Last updated date: 20th Jun 2024
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Hint: First of all, draw the diagram of cyclic quadrilateral by letting the angles and sides of the quadrilateral. Then use the angle sum property of a quadrilateral and property of angles in the same segment to prove the given statement.

Let $ABCD$be the given cyclic quadrilateral of a circle with centre at $O$.
Now construct a cyclic quadrilateral by joining the diagonal of the quadrilateral and naming the angles as shown in the figure:

We have to prove that sum of the angles of opposite sides are ${180^ \circ }$
i.e. $\angle BAD + \angle BCD = \pi$
$\angle ABC + \angle ADC = \pi$
By angle sum property of Quadrilateral
$\angle A + \angle B + \angle C + \angle D = 2\pi$…………………………………………………… (1)
By the property of angles in the same segment are equal
at chord $\overrightarrow {AB} {\text{, }}\angle e = \angle h$
at chord $\overrightarrow {BC} {\text{, }}\angle a = \angle g$
at chord $\overrightarrow {CD} {\text{, }}\angle b = \angle d$
at chord $\overrightarrow {AD,} {\text{ }}\angle c = \angle f$
From equation (1) we have
$\Rightarrow \angle A + \angle B + \angle C + \angle D = 2\pi \\ \Rightarrow \angle a + \angle b + \angle c + \angle d + \angle e + \angle f + \angle g + \angle h = 2\pi \\$
Using the above relations, we get
$\Rightarrow 2\left( {\angle a + \angle b + \angle e + \angle f} \right) = 2\pi \\ \Rightarrow \angle a + \angle b + \angle e + \angle f = \pi \\ \Rightarrow \left( {\angle a + \angle b} \right) + \left( {\angle e + \angle f} \right) = \pi \\$
From the figure clearly, we have
$\angle BAD + \angle BCD = \pi$
Similarly, we can write
$\angle ABC + \angle ADC = \pi$
Hence the sum of opposite angles of a Cyclic Quadrilateral is $\pi$.

Note: Angle sum property of quadrilateral means the sum of all angles in the quadrilateral is equal to $2\pi$. Always remember that the angles in the same segment of quadrilateral are equal. Remember the given statement as a property of cyclic quadrilaterals.