Question

From the following figure;

prove that:
(i) AB > BD
(ii) AC > CD
(iii) AB + AC > BC.

Answer 1

(i) ∠ADC + ∠ADB = 180°      ...[ BDC is a straight line ]
∠ADC = 90°                          ...[ Given ]
90° + ∠ADB = 180°
∠ADB = 90°                           ....(i)

In ΔADB,
∠ADB = 90°                           ....[ From (i) ]
∴ ∠B + ∠BAD = 90°
Therefore, ∠B and ∠BAD are both acute, that is less than 90°.
∴ AB > BD                                 ….(ii)[ Side opposite 90° angle is greater than the side opposite acute angle ]

(ii) In ΔADC,
∠ADB = 90°
∴ ∠C + ∠DAC = 90°
Therefore, ∠C and ∠DAC are both acute, which is less than 90°.
∴ AC > CD                       ...(iii)[ Side opposite 90° angle is greater than side opposite acute angle ]
Adding (ii) and (iii)
AB + AC > BD + CD
⇒ AB + AC > BC