The perpendicular from A on side BC of a triangle ABC intersects BC at D such that DB = 3CD (see figure). Prove that \(2{AB}^2\) = \(2{AC}^2\) + \({BC}^2\)

Solution :

We have : DB = 3CDtriangle

Now, BC = DB + CD

i.e. BC = 3CD + CD          [because BD = 3CD]

BC = 4CD

\(\therefore\)   CD = \(1\over 4\) BC  and  DB = 3CD = \(3\over 4\) BC             ……….(1)

Since triangle ABD is a right triangle, right angled at D, therefore, by Pythagoras Theorem, we have :

\({AB}^2\) = \({AD}^2\) + \({DB}^2\)             ……….(2)

In triangle ACD,

\(\angle\) D = 90 ,  \({AC}^2\) = \({AD}^2\) + \({CD}^2\)              ……..(3)

Subtracting (3) from (2), we get

\({AB}^2\) – \({AC}^2\) = \({DB}^2\) – \({CD}^2\)

\({AB}^2\) – \({AC}^2\) = \(({3\over 4}BC)^2\) – \(({1\over 4}BC)^2\)              (using (1))

\({AB}^2\) – \({AC}^2\) = \(({9\over 16} – {1\over 16})\)\({BC}^2\)

\({AB}^2\) – \({AC}^2\) = \({1\over 2}\)\({BC}^2\)

\(\implies\)  \(2{AB}^2\) = \(2{AC}^2\) + \({BC}^2\)

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