Question

In a triangle ABC, N is appoint on AC such that $BN\perp AC$. If ${BN}^2=AN\cdot NC$, prove that $\angle B=90.$

Answer 1

Let $\angle B={90}^{\circ }$ and T.Prove $\to {BN}^2=AN.NC$

If $\angle BAC=\theta$ In $\Delta ANB$ & $\Delta CNB$

$\tan \theta =\frac{BN}{AN},\tan \theta =\frac{CN}{BN}$

therefore from above, $\angle CBN=\theta$

On comparing, $\frac{BN}{AN}=\frac{CN}{BN}$

$\Rightarrow {BN}^2=AN\times CN$ (H.P)

So, $\angle B={90}^0$