If the median of Delta ABC through A is perpe | vedclass

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(C) Let $AD$ be the median through $A$ to the side $BC$. We are given that $AD \perp AB$,so $\angle DAB = 90^{\circ}$.Draw $CN \perp AB$ (produced). I

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$	an A + 2	an B = 0$

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(C) Let $AD$ be the median through $A$ to the side $BC$. We are given that $AD \perp AB$,so $\angle DAB = 90^{\circ}$.Draw $CN \perp AB$ (produced). In $\Delta BCN$,$AD \parallel CN$ because both are perpendicular to $AB$.Since $D$ is the midpoint of $BC$,by the intercept theorem,$A$ is the midpoint of $BN$,so $AB = AN$.In $\Delta ACN$,$\angle CAN = \alpha$. Then $\angle CAB = 180^{\circ} - \alpha$,so $	an A = 	an(180^{\circ} - \alpha) = -	an \alpha$.In $\Delta ACN$,$	an \alpha = \frac{CN}{AN}$.In $\Delta ABD$,$	an B = \frac{AD}{AB}$.Since $AD = \frac{1}{2}CN$ and $AB = AN$,we have $	an \alpha = \frac{2AD}{AB} = 2	an B$.Therefore,$	an A = -2	an B$,which implies $	an A + 2	an B = 0$.

If the median of $\Delta ABC$ through $A$ is perpendicular to $AB$,then

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