Question

If the non parallel sides of a trapezium are equal prove it is cyclic

Answer 1

Answer:

THIS IS YOUR ANSWER

Step-by-step explanation:

Given:

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

Construction:

DMDM and CNCN are perpendicular drawn on ABAB from DD and CC respectively.

To prove:

ABCDABCD is cyclic trapezium.

Proof:

In \triangle DAM△DAM and \triangle CBN△CBN,

AD = BCAD=BC ...Given

∠AMD = ∠BNC∠AMD=∠BNC ...Right angles

DM = CNDM=CN ...Distance between the parallel lines

\triangle DAM ≅ \triangle CBN△DAM≅△CBN by RHS congruence condition.

Now,

∠A = ∠B∠A=∠B ...by CPCT

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

⇒ ∠A + ∠C = 180°⇒∠A+∠C=180°

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

solution