Question

If the non-parallel sides of a trapezium are equal, prove that it is cyclic.​

Answer 1

Step-by-step explanation:

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$\huge\pink{\mid{\fbox{\tt{QųEsTɪOη}}\mid}}$

If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

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$\huge{\underline{\mathrm{Answer}}}$

ABCD is a trapezium where non-parallel sides AD and BC are equal.

$\huge{\underline{\mathrm{Construction:}}}$

∴DM and CN are perpendicular drawn on AB from D and C, respectively.

$\huge{\underline{\mathrm{To\: prove:}}}$

ABCD is cyclic trapezium.

$\huge{\underline{\mathrm{Proof:}}}$

In △DAM and △CBN,

∴AD = BC ... [Given]

∠AMD =∠BNC ...[Right angles]

∴DM = CN ...[Distance between the parallel lines]

∴Therefore, △DAM≅△CBN by RHS congruence condition.

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Now, ∠A =∠B ...[by CPCT]

Also, ∠B+∠C=180° ....[Sum of the co-interior angles]

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${\boxed{⇒∠A+∠C = 180° }}$

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Thus, ABCD is a cyclic quadrilateral as the sum of the pair of opposite angles is 180°.

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$\huge{\underline{\mathrm{Note}}}$

Please refer to the attachment also ❤️

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