if the non-parrallel sides of a trapezium are equal , prove that it is cyclic?
Answers
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Step-by-step explanation:
Answer
Given:
ABCD is a trapezium where non-parallel sides AD and BC are equal.
Construction:
DM and CN are perpendicular drawn on AB from D and C, respectively.
To prove:
ABCD is cyclic trapezium.
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.
Now, ∠A=∠B ...[by CPCT]
Also, ∠B+∠C=180° ....[Sum of the co-interior angles]
⇒∠A+∠C=180°.
Thus, ABCD is a cyclic quadrilateral as the sum of the pair of opposite angles is 180°.
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