Question

solve this problem jalfi​

Answer 1

so here is your answer...

in ∆ABC

AD is median

so, BD=CD

Draw AE perpendicular to BC

since, angle AED=90°

therefore in ∆AED

Angle ADE <90° and angle ADB>90°

∆ABD is obtuse angled and ∆ACD is acute angled

∆ABD is obtuse at angle D and AE is perpendicular to BD produced.

therefore,

AB^2 =AD^2+BD^2+2BD×DE _______(1)
∆ACD is acute angled at D and AE is perpendicular to BD produce
therefore,
AC^2=AD^2+BD^2-2BD ×DE________(2) [since BD=CD]
Adding (1) &(2)
AB^2+AC^2= 2AD^2+ 2BD^2
AB^2+AC^2=2(AD^2+BD^2)
hence, proved...

Hope it will be helpful for you..

Answer 2

Please see the attachment