Question

if the non parallel side software a parallelogram are equal prove that it is cyclic
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Answer 1

Here is your 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

△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 sum of the pair of opposite angles is 180°.

Step-by-step explanation:

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Answer 2

Answer:

Heya friend,!!

Given: ABCD is parallelogram a where AB||CD and AD = BC

To prove: ABCD is cyclic.

Construction: Draw DL⊥AB and CM⊥AB.

Proof: In ΔALD and ΔBMC,

AD = BC (given)

DL = CM (distance between parallel sides)

∠ALD = ∠BMC (90°)

ΔALD ≅ ΔBMC (RHS congruence criterion)

⇒ ∠DAL = ∠CBM (C.P.C.T) (1)

Since AB||CD,

∠DAL + ∠ADC = 180° (sum of adjacent interior angles is supplementary)

⇒ ∠CBM + ∠ADC = 180° (from (1))

⇒ ABCD is a cyclic parallelogram (Sum of opposite angles is supplementary)

Hope it helps u