Question

If in triangle ABC ,AD is a median and AM is perpendicular to BC ,then prove that
AB2=AD2-BC×DM + 1/4BC2

Answer 1

Answer:

Step-by-step explanation:

AB2=AM2+BM2          -{1}

AD2=AM2+MD2

AM2=AD2-MD2          -{2}

also, BM= BC-(DM+DC)

BUT DC=1/2 BC

therefore, BM=BC-(DM+1/2BC)       -{3}

now using equ.2 and 3 in equ.1

AB2=AD2-MD2+[BC-(DM+1/2BC)2

AB2=AD2-MD2+[BC-DM-1/2BC]2

AB2=AD2-MD2+[BC/2-DM]2

by solving this it will prove that AB2= AD2-BC*DM+1/4BC