Question

2.
Prove that any outer angle of a cyclic quad-
rilateral is equal to the inner angle at the
opposite vertex.​

Answer 1

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✤ Required Proof:

✒ GiveN:

• A cyclic quadrilateral.

✒ To prove:

• Outer angle of a cyclic quadrilateral is equal to the inner angle at opposite vertex !!

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✤ How to Solve?

For proving the above statement, we need to know a theoram based on Cyclic quadrilaterals( The quadrilateral inside a circle, whose vertices touched the circumference)

$\large{ \bf{Theoram:}}$

➤ The sum of opposite angles in a cyclic quadrilateral is supplementary i.e. they add upto 180°

So, Let's use this theoram, to prove this statement.

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✤ Solution:

✒ Refer to the attachment.....

According to theoram,

⇛ ∠CBA + ∠CDA = 180°...........(1)

[ Sum of opposite angles = 180°]

Now in line AE,

⇛ ∠CBA + ∠CBE = 180°...........(2)

[ These angles add upto 180° as they form a straight line, hence they are linear pair.]

From equation, (1) and (2)

⇛ ∠CBA + ∠CDA = ∠CBA + ∠CBE

⇛∠CDA = ∠CBE ( ∠CBA cancels both sides)

Here,

• ∠CDA = Angle at opposite to ∠CBA
• ∠CBE = Outer/Exterior angle of ∠CBA

✒ And, We proved that both of these angles are equal, Hence, Proved !

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Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{ \angle{CBE } = \angle{ADC}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}}$

$\tt: \implies a \: cyclic \: quadrilateral \: \angle{ABCD} \: whose \: AB \: prodused \: to \: a \: point \: E$

$\red{\underline \bold{To \: Prove :}}$

$\tt: \implies external \: \angle{CBE} = \angle{ADC}$

• Given Question

$\tt: \implies Prove \: \angle{ABC} + \angle{CBE} = 180°$

$\tt: \implies \angle{ABC} + \angle{ADC} = 180°$

$\tt: \implies \angle{ABC} + \angle{CBE} = \angle{ABC} + \angle{ADC}</strong><strong>$

$\tt: \implies \angle{ABC} + \angle{CBE} = \angle{ABC} + \angle{ADC}$

$\green{\tt{\therefore{ \angle{CBE } = \angle{ADC}}}}$

$\tt\: hence \: proved$