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प्रश्न
From the following figure, prove that: AB > CD.
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उत्तर
In ΔABC,
AB = AC ...[ Given ]
∴ ∠ACB = ∠B ...[ angles opposite to equal sides are equal ]
∠B = 70° ...[ Given ]
⇒ ∠ACB = 70° ...(i)
Now,
∠ACB +∠ACD = 180°...[ BCD is a straight line]
⇒ 70° + ∠ACD = 180°
⇒ ∠ACD = 110° ...(ii)
In ΔACD,
∠CAD + ∠ACD + ∠D = 180°
⇒ ∠CAD + 110° + ∠D = 180° ...[ From (ii) ]
⇒ ∠CAD + ∠D = 70°
But ∠D = 40° ...[ Given ]
⇒ ∠CAD + 40°= 70°
⇒ ∠CAD = 30° …(iii)
In ΔACD,
∠ACD = 110° ...[ From (ii) ]
∠CAD = 30° ...[ From (iii) ]
∠D = 40° ...[ Given ]
∴ D > ∠CAD
⇒ AC > CD ....[Greater angle has greater side opposite to it]
Also,
AB = AC ...[ Given ]
Therefore, AB > CD.
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