Consider the following statements: 1) The perimeter of a triangle is greater than the sum of its three medinas. 2) In any triangle ABC, if D is any point on BC, then AB + BC + CA > 2AD. Which of the above statements is/are correct?
Options
A. 1 only
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2
Correct Answer
Both 1 and 2
Explanation
Let ABC be the triangle and D. E and F are midpoints of BC, CA and AB respectively.
Recall that the sum of two sides of a triangle is greater than twice the median bisecting the third side,(Theorem to be remembered)
Hence in ?ABD, AD is a median
?
AB + AC > 2(AD)
Similarly, we get
BC + AC > 2CF
BC + AB > 2BE
On adding the above inequations, we get
(AB + AC) + (BC + AC) + (BC + AB )> 2AD + 2CD + 2BE
2(AB + BC + AC) > 2(AD + BE + CF)
?
AB + BC + AC > AD + BE +CF
2.
To prove: AB + BC + CA > 2AD
Construction: AD is joined
Proof: In triangle ABD,
AB + BD > AD [because, the sum of any two sides of a triangle is always greater than the
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