Question

In ΔABC, AB > AC, D is the mid-point of BC . AM ⊥ BC such that B-M-C. Prove that AB²-AC²=2BC.DM.

Answer 1

In ΔABC, AB > AC, D is the mid-point of BC . AM ⊥ BC such that B-M-C.

as AM ⊥ BC,
ΔACM and ΔABM are right triangle.
By using Pythagoras Theorem, we get,
AC² = CM² + AM² ...... (1)
Also, AB² = BM² + AM² ...... (2)
Eliminating AM² from equations (1) and (2), we get,
AC² – CM² = AB² – BM²
⇒ AC² – AB² = CM² – BM²
By using the identity a² – b² = (a + b)(a – b) on above equation,
⇒ AC² – AB² = (CM + BM)(CM – BM)
⇒ AC² – AB² = BC(CM – (BC – CM))
⇒ AB² – AC² = BC(2CM – BC)
As D is mid point of BC,
⇒ AB² – AC² = BC(2CM – 2BD)
⇒ AB² – AC² = 2BC(CM – BD)
⇒  AB² – AC² = 2BC(CM – CD)
⇒ AB² – AC² = 2BC(–DM)
⇒ AB² – AC² = 2BC.DM
hence proved.

Answer 2

In ∆ABC ,

AB > AC , D is the midpoint of BC.

BD = DC --( 1 )

AM is perpendicular to BC .

i ) In ∆AMB , <M = 90°

AB² = BM² + AM² --( 2 )

[ By Phythogarian theorem ]

ii ) In ∆AMC , <M = 90°

AC² = MC² + AM² ---( 3 )

Subtract equation ( 3 ) from ( 2 ) ,

we get

AB² - AC² = BM² - MC²

= ( BM + MC )( BM - MC )

= BC [(BD+DM )-(DC - DM)]

= BC[( BD + DM)-(BD - DM)]

= BC[BD + DM - BD + DM ]

= BC ( 2DM )

= 2BC × DM

Therefore ,

AB² - AC² = 2BC× DM

I hope this helps you.

: )