$ABC$ is a triangle right-angled at $C$. $A$ line through the mid-point $M$ of hypotenuse $AB$ and parallel to $BC$ intersects $AC$ at $D$. Show that $MD \perp AC$.
(N/A) We have a triangle $ABC$ such that $\angle C = 90^{\circ}$. $M$ is the mid-point of $AB$ and $MD \parallel BC$.
To prove: $MD \perp AC$.
Since $MD \parallel BC$ and $AC$ is a transversal,
$\therefore \angle MDA = \angle BCA$ [Corresponding angles].
But $\angle BCA = 90^{\circ}$ [Given].
$\therefore \angle MDA = 90^{\circ}$.
$\Rightarrow MD \perp AC$.
To prove: $MD \perp AC$.
Since $MD \parallel BC$ and $AC$ is a transversal,
$\therefore \angle MDA = \angle BCA$ [Corresponding angles].
But $\angle BCA = 90^{\circ}$ [Given].
$\therefore \angle MDA = 90^{\circ}$.
$\Rightarrow MD \perp AC$.
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