The perpendicular \(AD\) on the base \(BC\) of a \(\Delta ABC\) intersects \(BC\) at \(D\), such that \(BD\) \(=\) \(3\) \(DC\). Prove that \(2\) \(AB^2 = 2AC^2 + BC^2\).
Proof:
From the figure we have, \(BC\) \(=\) \(BD\) \(+\) \(DC\).
\(\Rightarrow\) \(BC\) \(=\) [Given]
\(\Rightarrow\) \(=\) \(\frac{BC}{4}\) \(……. (1)\)
Substitute equation \((1)\) in \(BD\) \(=\) \(3\) \(DC\).
\(\Rightarrow\) \(BD\) \(=\) \(3\) \(\times\) \(\frac{BC}{4}\)
\(\Rightarrow\) \(BD\) \(=\) \(\frac{3BC}{4}\) \(……. (2)\)
\(\Rightarrow\) \(BD\) \(=\) \(3\) \(\times\) \(\frac{BC}{4}\)
\(\Rightarrow\) \(BD\) \(=\) \(\frac{3BC}{4}\) \(……. (2)\)
Consider the right triangle \(ADB\).
We have, \(…… (3)\) [By Pythagoras theorem]
Consider the right triangle \(ADC\).
We have, \(…… (4)\) [By the Pythagoras theorem]
Subtract equation \((4)\) from \((3)\).
\(AB^2 - AC^2 =\)
Substitute equation \((1)\) and \((2)\) in the above equation.
\(AB^2 - AC^2 =\)
\(2AB^2 - 2AC^2 = BC^2\)
\(2AB^2 = 2AC^2 + BC^2\)
\(2AB^2 = 2AC^2 + BC^2\)
Hence, proved.