Page 2 - EXERCISE 7.3
P. 2
MAHEISHANG CLASS IX BOSEM
BOOKS AND TUTORIAL
lets assumed in ABC
we will used exterior angle property
In AQB,
AQC ABQ ex angle is greater than one of its opp angle
. i e AB>AQ eqn 1
In BRC,
le
BRA> BCR ex angle is greater than one of its opp ang
i.e BC>BR eqn 2
In CPA
CPB> CAP ex angle is greater than one of its opp angle
i.e CA>CP eqn 3
adding eqn 1 ,eqn 2 and eqn 3 ,we get
AB+BC+CA>AQ+BR+CP h/p
Q4.Show that the difference of any two sides of a triangle is less than the third side.
soln:
construction : C on AD and joined BC
to prove: AB-BD<AD or BD-AD<AB or AD-AB<BD
proof:
5 2 exterior angle is greater than one of the opp angle
i.e AC<BD eqn 1
6< 5 exterior angle is greater than one of the opp angle
. i e CD<AB eqn 2
adding eqn 1 and eqn 2,we get
AC+CD<BD+AB
AD BD AB
AD AB BD h/p
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