# Q14   The perpendicular from A on side BC of a $\Delta$ ABC intersects BC at D such that DB = 3 CD           (see Fig. 6.55). Prove that $2 AB^2 = 2 AC^2 + BC^2.$

In $\triangle$ ACD, by Pythagoras theorem,

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

$AC^2-DC^2=AD^2..................1$

In $\triangle$ ABD, by Pythagoras theorem,

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

$AB^2-BD^2=AD^2.................2$

From 1 and 2, we get

$AC^2-CD^2=AB^2-DB^2..................3$

Given : 3DC=DB, so

$CD=\frac{BC}{4}\, \, and\, \, BD=\frac{3BC}{4}........................4$

From 3 and 4, we get

$AC^2-(\frac{BC}{4})^2=AB^2-(\frac{3BC}{4})^2$

$AC^2-(\frac{BC^2}{16})=AB^2-(\frac{9BC^2}{16})$

$16AC^2-BC^2=16AB^2- 9BC^2$

$16AC^2=16AB^2- 8BC^2$

$\Rightarrow 2AC^2=2AB^2- BC^2$

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

Hence proved

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