Question

In traingle ABC, BD is perpendicular on AC and BC^2 = 2 AC.CD. then prove that AB=AC

Answer 1

Given that BD is perpendicular to AC and BC2 = 2 AC.CD

=BD2 + CD2 =2AC.CD(Using pythogoras theorem in triangle BCD)

Again, in triangle ABD Using pythogoras theorem,

AB2 =BD2 + AD2

BD2= AB2 - AD2

So, AB2 - AD2 +CD2=2AC.CD

AB2 – (AC  - CD)2 + CD2=2AC.CD

AB2 – AC2  - CD2  + 2AC.CD + CD2=2AC.CD

AB2 – AC2  = 0

AB2 = AC2  

AB=AC

Answer 2

BC^2=2AC.AD (GIVEN)
BC^2=BD^2+CD^2 (SINCE B.D.c) IS A RIGHT ANGLED TRIANGLE)
BC^2=(AC-AD)^2+AB^2-AD^2
A.T.Q
2AC.CD=AC^2-2AC.AD+AB^2 =
2AC.CD-2AC.AD=AC^2+AB^2 =
2AC^2 =AC^2+AB^2
=AC^2=AB^2
AC=AB
HENCE PROVED