$A B C$ is a triangle right angled at $C$ . A line through the midpoint $M$ of hypotenuse $A B$ and Parallel to $B C$ intersects $A C$ at $D$ . Show that
$D$ is the midpoint of $A C$

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Solution

In $Δ A M D$ And $Δ A B C$
$∠ A$ is common;

$∠ B = ∠ M$ (corresponding angles)
$∠ C = ∠ D$ (BC is parallel to MD and BC is perpendicular to AC therefore, MD is also perpendicular to AC)..........(i)
$Δ A M D ≅ Δ A B C$ (by AAA)

$∴ \frac{A M}{A B} = \frac{A D}{A C} = \frac{1}{2}$
$\Rightarrow \frac{A D}{A D + C D} = \frac{1}{2}$

$\Rightarrow 2 A D = A D + C D$

$\Rightarrow A D = C D$

Hence the D is mid point of AC

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