Question

From the following figure, prove that: AB > CD.

Answer 1

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.