Question

In given fig. Angle B smaller than angle A. Angle C smaller than angle D. Show that AD is smaller than BC.

Answer 1

The image of the question is attached below.

In ΔAOB,

Angle B < Angle A

That is $\angle B < \angle A$

In any triangle, the side opposite to the greater angle is longer.

⇒ AO < BO – – – – (1)

In ΔCOD,

Angle C < Angle D

That is $\angle C < \angle D$

In any triangle, the side opposite to the greater angle is longer.

⇒ OD < OC  – – – – (2)

Adding equation (1) and equation (2), we get

⇒ AO + OD < BO + OC

⇒ AD < BC

So, AD is smaller than BC.

Hence proved.

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Answer 2

Answer:

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Step-by-step explanation:

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