Question

If D is the mid point of AB in triangle ABC then i) BC +CA > 2CD ii) AB+BC >2CD iii) AB+BC+CA > 2AD iv) AB+AC > 2AD

Answer 1

Answer:

From the figure ,

i ) In a triangle ABC , AD is the median

drawn on the side BC is produced

to E such that AD = ED then ABCD

is a parallelogram .

AE = AD + DE

=> AE = 2AD ---( 1 )

AC = BC ---( 2 )

[ Opposite sides of parallelogram ]

ii ) The sum of any two sides of a

triangle is greater than the third side .

Now ,

In ∆ABE ,

AB + BE > AE

=> AB + AC > 2AD [ from ( 1 ) & ( 2 ) ]

Answer 2

Answer:

I) This can be proved by making a small construction. Extend AD to E such that AD = DE. Join BE and CE. So, AE = 2AD ---------- (1)

ii) In the quadrilateral, ABEC, the diagonal AE and BC bisect each other at D. [Since by construction AD = DE and as given AD is the median to BC; so D is the midpoint of BC]

Since diagonals bisect each other, the quadrilateral ABEC is a parallelogram.

In a parallelogram opposite sides are equal; so BE = AC --------- (2)

iii) In triangle ABE, AB + BE > AE [Sum of any two sides of a triangle is greater than

the third side]

So from (1) & (2)

AB + AC > 2AD

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Step-by-step explanation:

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