In the given figure, ABCD is a parallelogram. If AM bisects $∠ A$ and DM bisects $∠ D$ , then $∠ A M D$ = ____.

60°

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75°

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90°

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120°

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Solution

The correct option is C
90°

Let $∠ A = 2 x$

Therefore, $∠ D$ = 180° - 2x° (adjacent angles of a parallelogram are supplementary)

$∠ D A M$ = $∠ M A B$ = x° (AM bisects $∠ A$ ) ……… (i)

$∠ A D M$ = $∠ M D C$
= $\frac{180 ° - 2 x °}{2}$ (AM bisects $∠ D$ )
= 90° - x° ……… (ii)

In AMD,
$∠ D A M$ + $∠ A D M$ + $∠ A M D =$ 180° (sum of angles of a triangle)
x° + 90 - x + $∠ A M D$ = 180°
$∠ A M D$ = 90°

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∠ A

∠ D

∠ A M D = 90^{\circ}

ABCD is a parallelogram, M is the mid-point of BD and BM bisects $∠ B$ . Then $∠ A M B =$

In the given figure, ABCD is a parallelogram, M is the midpoint of BD and BD bisects $∠ B$ as well as $∠ D$ . Then, $∠ A M B =$ ?

(a) $45^{\circ}$

(b) $60^{\circ}$

(d) $30^{\circ}$

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