Question

In the Given Figure angle B< angle A and angle C < angle D show that AD < BC

Answer 1

#ANSWER...!

__________________________

INEQUALITIES IN TRIANGLE....

__________________________

Show that -: AD < BC..

Given ,

Angle B < Angle A

So , AO < BO

(Side opposite of greater angle is longest and smaller angle is shorter)

Angle C < Angle D

So , DO < CO

(Side opposite of greater angle is longest and smaller angle is shorter)

Now ,

AO + DO < BO + CO

Since ,

AO + DO = AD
BO + CO = BC

Therefore ,

AD < BC....

-- Proved --