Math, asked by bmathew1998, 1 year ago

In triangle ABC ,if AD is the median, then show that AB^2+AC^2=2(AD^2+BD^2)

Answers

Answered by shadowsabers03

Draw an altitude AM from A to BC at M.

Here, BD = CD.

Assume that AM is at left of AD. See the figure.

So that,

AB^2 = BM^2 + AM ^2

=> AB^2 = (BD - DM)^2 + AM^2

=> AB^2 = BD^2 - 2•BD•DM + DM ^2 + AM ^2

=> AB^2 = BD^2 - 2•BD•DM + AD^2 --> (1)

And,

AC^2 = CM^2 + AM^2

=> AC^2 = (CD + DM)^2 + AM^2

=> AC^2 = CD^2 + 2•CD•DM + DM^2 + AM^2

=> AC^2 = CD^2 + 2•CD•DM + AD^2 --> (2)

(1) + (2)

AB^2 + AC^2 = BD^2 - 2•BD•DM + AD^2 +CD^2 + 2•CD•DM + AD^2

=> AB^2 + AC^2 = 2AD^2 + CD ^2 + 2DM(CD - BD) + BD^2

=> AB^2 + AC^2 = 2AD^2 + BD^2 + 2DM(BD - BD) + BD^2 [Because BD = CD]

=> AB^2 + AC^2 = 2AD^2 + BD^2 + 0 + BD^2

=> AB^2 + AC^2 = 2AD^2 + 2BD^2

=> AB^2 + AC^2 = 2(AD^2 + BD^2)

Hence proved!

Plz mark it as the brainliest.

Thank you. :-))

Attachments:

shadowsabers03: Thank you for marking my answer as the brainliest.

Answered by sshazu5856

Answer:

by pythagoras property,

Step-by-step explanation:

Attachments:

bmathew1998: nope

shadowsabers03: AD is median, not altitude!!!

sshazu5856: even if it is median it divides into two equal sides...i even checked in google.....it says

sshazu5856: In the case of isosceles and equilateral triangles, a median bisects any angle at a vertex whose two adjacent sides are equal in length. This is what google says......so its correct

sshazu5856: oops kk tht was not isoceles or euilateral kk kk i understand

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