In the adjacent figure, D is any point on the side BC of ∆ABC. Show that AB + BC + CA > 2AD
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Answered by
8
Answer:
Step-by-step explanation:
In triangle ABD
AB+BD>AD
Similarly in triangle ADC
AC+CD>AD
This implies that
AB+BD+AC+CD>AD+AD
ie
AB+BC+CA>2AD
Answered by
12
Answer:
In ABD , By Inequality property of triangle
AB + BD > AD ----------(1)
And In ACD ,By Inequality property of triangle
DC + AC > AD ---------(2)
On Adding Eq (1) and Eq (2)
AB+BD+DC+AC> AD+AD
AB + BC + AC > 2 AD [Given BD+DC = BC]
AB + BC + AC > 2 AD
hence Proved
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