Show that the sum of any two sides of a triangle is greater than the third side
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Answer:
Given: In triangle ABC, AD is the median drawn from A to BC.
To prove: AB + AC > AD
Construction: Produce AD to E so that DE = AD, Join BE.
Proof:
Now in ADC and EDB,
AD = DE (by const)
DC = BD(as D is mid-point)
ADC = EDB (vertically opp. s)
Therefore,
In ABE, ADC EDB(by SAS)
This gives, BE = AC.
AB + BE > AE
AB + AC > 2AD (AD = DE and BE = AC)
Hence the sum of any two sides of a triangle is greater than the median drawn to the third side
Step-by-step explanation:
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