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Trigonometric Ratios

ABC is a tria...

Question

ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallels to BC intersect AC at D. Show that CM=MA= $\frac{1}{2} A B$

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Solution

To show : $C M = M A = \frac{1}{2} A B$

In $△ A M D$ and $△ C M D$

$A D = D C$ ( $D$ is midpoint)

$∠ A D M = ∠ C D M$

$D M = D M$ (common)

$△ A M D =≅ △ C M D$ ( $S A S$ congruence rule )

$∴ A M = C M (C P C T) - - - - - - - - - - - - - (1)$

$A M = \frac{1}{2} A B (M i s m i d p o i n t) - - - - - - - - - - (2)$

from eq $(1)$ and $(2)$

$C M = A M = \frac{1}{2} A B$

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