Question

41. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
OR
Construct ABC in which base AB = 5 cm, ZA = 30° and AC- BC = 2.5 cm.​

Answer 1

Step-by-step explanation:

$Given:ABCD \:  is \: a \: \: trapezium \: where \: non-parallel \: sides \:  AD \:  and \:  BC \:  are \: equal.$

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°.

Answer 2

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