Question

in the given figure, AC>AB, and D is a point on AC such that AB=Ad. prove that BC>CD

Answer 1

AB=AD⇒∠ABD=∠ADB             (side opp. to equal ∠s are equal)
∠B=∠ABD+∠DBC
∠DBC=∠B-∠ABD=90°-∠ABD     (clearly,from the figure ∠B=90°)

∠ADB+∠BDC=180°
∠BDC=180°-∠ADB
=180°-∠ABD

For sake of convenience take ∠ABD as any number you wish,here we take it as 1
∠DBC=90°-1=89,∠BDC=180-1=179
Then,∠BDC>∠DBC
⇒BC>CD                   (side opp. to the greater angle is longer)
∴BC>CD

Answer 2

Answer:

In triangle ABC

AB+BC > AC

AB+BC > AD+DC

AB+BC > AB+DC [ ∵ AB=AD (given) ]

∴ BC > DC

Hope it helps u