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प्रश्न
In the following figure ; AC = CD; ∠BAD = 110o and ∠ACB = 74o.
Prove that: BC > CD.
उत्तर
∠ACB = 74° ...(i)[ Given ]
∠ACB + ∠ACD = 180° ....[ BCD is a straight line ]
⇒ 74° + ∠ACD = 180°
⇒ ∠ACD = 106° …..(ii)
In ΔACD,
∠ACD + ∠ADC+ ∠CAD = 180°
Given that AC = CD
⇒ ∠ADC= ∠CAD
⇒ 106° + ∠CAD + ∠CAD = 180° ....[From (ii)]
⇒ 2∠CAD = 74°
⇒ ∠CAD = 37° = ∠ADC ...(iii)
Now,
∠BAD = 110° ....[Given]
∠BAC + ∠CAD = 110°
∠BAC + 37° = 110°
∠ BAC = 73° ….(iv)
In ABC,
∠ B + ∠BAC + ∠ACB = 180°
∠B + 73° + 74° = 180° ...[From (i) and (iv)]
∠B + 147°= 180°
∠B = 33° …..(v)
∴ ∠BAC > ∠B ...[ From (iv) and (v)]
⇒ BC > AC
But,
AC = CD ...[ Given ]
⇒ BC > CD
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