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The perpendicular from A on the side BC of a $\bigtriangleup ABC$ intersects BC at D, such that $D\!B = 3 C\!D$ . Prove that $2 AB^2 = 2 AC^2 + BC^2$ .

Answers (1)

Given $AD\perp BC$ . Also $D\!B = 3 C\!D$

Let $BC = x$

$CD+BD = x$

$CD+3CD = x$

$C\!D = \frac{x}{4}$

Putting the value of CD in given equation :

$DB=3CD$

$DB=\frac{3x}{4}$

In $\bigtriangleup ADC$ , by using pythagoras theorem

$AC^2=DC^2+AD^2$

$AC^2=\left ( \frac{x}{4} \right )^2+AD^2$

$AC^2=\frac{x^2}{16} +AD^2 \;\;\;\;\; -(1)$

In $\bigtriangleup ABD$ ,

$AB^2 = BD^2 + AD^2$

$\Rightarrow AB^2 = \left ( \frac{3x}{4} \right )^2 + AD^2$

$\Rightarrow AB^2 = \frac{9x^2}{16} + AD^2 \;\;\;\;\;\; - (2)$

Substracting eqn (2) by eqn (1) :

$AC^2 - AB^2 = AD^2 + \frac{x^2}{16} - \left ( AD^2 + \frac{9x^2}{16} \right )$

$AC^2 - AB^2 = \frac{-8x^2}{16}$

$AC^2 - AB^2 = \frac{-x^2}{2}$

$2AC^2 - 2AB^2 = -x^2$

$2AC^2 - 2AB^2 = -BC^2 \;\;\;\;\;\; \left ( \because BC = x \right )$

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

Hence proved.

Posted by

Ravindra Pindel

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