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

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.

      

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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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