Question

If in the triangle ABC, AD is a median and AM is perpendicular to BC then prove that AB^2 + AC^2 = 2 (AD^2 + BD^2). No diagram provided

Answer 1

$since \: ad \: is \: the \: median \: of \: triangle \: abc \\ bd = cd \\ then \: in \: triangle \: abd \: \\ by \: pythagoras \: theorem \\ ( {ab})^{2} = ( {ad})^{2} + ( {bd})^{2} ..............(i) \\ in \: triangle \: adc \\ by \: pythagoras \: theorem \\ ({ac})^{2} = ( {ad})^{2} + ( {cd})^{2} \\ ( {ac})^{2} = ( {ad})^{2} + ( {bc})^{2} ...........(ii) \\ adding \: eq \: (i) \: and \: (ii) \\ ( {ab})^{2} +( {ac})^{2} = ( {ad})^{2} + ( {bd})^{2} + ( {ad})^{2} + ( {bd})^{2} \\ ( {ab})^{2} + ( {ac})^{2} = 2( {ad)}^{2} + 2( {bd})^{2} \\ ( {ab})^{2} + ( {ac})^{2} = 2(( {ad)}^{2} + ( {bd})^{2})$
Heyy friend your answer is here

Answer 2

plz refer to this attachment......

Hope it helps you ........

correct answer........