In the given figure, AC >AB, and D is a point on AC such that AB =AD. prove that BC >CD
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in triangle abc and triangle bad
ac>ab->i
ab=ad and ac= bc (c and d is the mid points of line ab)
form I
ac>ab
bc>ad->2
but ad = cd (d is mid point of ac)
from 2
bc>cd
ac>ab->i
ab=ad and ac= bc (c and d is the mid points of line ab)
form I
ac>ab
bc>ad->2
but ad = cd (d is mid point of ac)
from 2
bc>cd
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