Question

In a right triangle abc right angled at c if d is the midpoint of bc prove that bc2=4(ad2-ac2)

Answer 1

Consider the right triangle ABC, right angled at C.

We have to prove: $BC^2 = 4(AD^2 - AC^2)$

Consider triangle ABC,

by Pythagoras theorem, we get

$(AB)^2 = (AC)^2 + (BC)^2$

Consider triangle ACD,

by Pythagoras theorem, we get

$(AD)^2 = (AC)^2 + (CD)^2$ (Equation 1)

Since, D is the midpoint of BC.

CD=DB and $CD=\frac{CB}{2}$

Substituting the value of CD in equation 1, we get

$(AD)^2 = (AC)^2 + (\frac{BC}{2})^2$

$(AD)^2 = (AC)^2 + \frac{BC^2}{4}$

$(AD)^2 = \frac{4(AC)^2 + (BC)^2}{4}$

$4(AD)^2 = {4(AC)^2 + (BC)^2}$

$(BC)^2 = 4(AD)^2 - 4 (AC)^2$

$(BC)^2 = 4[(AD)^2 - (AC)^2]$

Hence, proved.

Answer 2

Answer:

AC2 = BC2 + AB2. ... Triangle ABC is right- angled at B and D is the mid-point of BC. Prove ... AD2 = BD2 + AB2 ... AC squared equals 4 AD squared minus 4 AB squared plus AB squared