$\Delta ABC$ 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 $D$ is the mid-point of $AC$.
(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 that $D$ is the mid-point of $AC$.
In $\Delta ACB$,we have:
$M$ as the mid-point of $AB$. [Given]
$MD \parallel BC$. [Given]
Therefore,using the converse of the mid-point theorem,$D$ is the mid-point of $AC$.
To prove that $D$ is the mid-point of $AC$.
In $\Delta ACB$,we have:
$M$ as the mid-point of $AB$. [Given]
$MD \parallel BC$. [Given]
Therefore,using the converse of the mid-point theorem,$D$ is the mid-point of $AC$.
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