Question

AM is a median of a triangle ABC. Is AB +BC +CA + >2 AM

Answer 1

$\huge\red{\boxed{\bold{Solution}}}$

AM is a median.

AM is a median. So, BM = CM

Construction: extend AM to D such that AM = MD

ABCD is a parallelogram as diagonals are bisecting each other.

Since, AB + BM > AM — [eqn 1]

( the sum of two sides of a triangle > 3rd side)

& BD + BM > MD — [eqn 2]

( the sum of two sides of a triangle > 3rd side)

NOW BY ADDING 1 & 2 WE GET

AB + BD + 2BM > AM + MD — [eqn 3]

But BD = AC ( opposite sides of parallelogram)

& 2BM = BC

& AM = MD

so, eqn 3 becomes

AB + AC + BC = 2AM

[PROVED]

$\huge\mathbb\blue{THANK\:YUH!}$

Answer 2

AM is a median.

AM is a median. So, BM = CM

Construction: extend AM to D such that AM = MD

ABCD is a parallelogram as diagonals are bisecting each other.

Since, AB + BM > AM — [eqn 1]

( the sum of two sides of a triangle > 3rd side)

& BD + BM > MD — [eqn 2]

( the sum of two sides of a triangle > 3rd side)

NOW BY ADDING 1 & 2 WE GET

AB + BD + 2BM > AM + MD — [eqn 3]

But BD = AC ( opposite sides of parallelogram)

& 2BM = BC

& AM = MD

so, eqn 3 becomes

AB + AC + BC = 2AM

[PROVED]

$\huge\mathbb\blue{THANK\:YUH!}$