If the median of $△ A B C$ through A is perpendicular to AB prove that $t a n A + 2 t a n B = 0$ .

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Solution

We are given
BD= DC and $∠ D A B = 90^{0}$
Draw CN $⊥$ to BA produced
Then in $△ B C N$ we have
$D A = \frac{1}{2} C N = y$ and AB=AN =z.
$∴ t a n A = t a n (π - ∠ C A N)$
= $- t a n (∠ C A N) = - \frac{C N}{A N} = - \frac{2 y}{z}$
or $t a n A = - (2 A D / A B) = - 2 t a n B$
or $t a n A + 2 t a n B = 0$

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