Question

In Fig. 11.53, AB is the longest side and DC is the shortest side of a quadrilateral
ABCD. Prove that ∠C > ∠A and ∠D> ∠B. [Hint : Join AC and BD].

Answer 1

Answer:Given:

In quadrilateral ABCD, AB smallest & CD is longest sides.

To Prove: ∠A>∠C

& ∠B>∠D

Construction: Join AC.

Mark the angles as shown in the figure..

Proof:

In △ABC , AB is the shortest side.

BC > AB

∠2>∠4 …(i)

[Angle opposite to longer side is greater]

In △ADC , CD is the longest side

CD > AD

∠1>∠3 …(ii)

[Angle opposite to longer side is greater]

Adding (i) and (ii), we have

∠2+∠1>∠4+∠3

⇒∠A>∠C

Similarly, by joining BD, we can prove that

∠B>∠D

Please mark as the brainliest answer..................................

Answer 2

Answer:

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Step-by-step explanation:

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