Question

In ∆ABC, if AD is the median, then show that AB² + AC² = (AD² + BD²)

Answer 1

Answer:

given BD=DC,

BD=AB

Explanation:

Both of square root then both square is equal to answer

Answer 2

AB2 + AC2 = 2[AD2 + BD2]

Explanation:

Given: AD is a median in ABC.

To prove: AB2 + AC2 = 2[AD2 + BD2]

AE perpendicular to BC.

In right-angled triangles AEB and AEC, using Pythagoras theorem we get:

AB^2 + AC^2 = BE^2 + AE^2 + EC^2 + AE^2

 = 2AE^2 + (BD - ED)^2 + (ED + DC)^2

 = 2AE^2 + 2ED^2 + BD^2 + DC^2

AB^2 + AC^2 = 2AE^2 + 2ED^2 + 2BD^2 [since BD = DC]

= 2 (AE^2 + ED^2 + BD^2)

= 2 (AD^2 + BD^2)

Hence proved.