# 1) x and y are respectively the mid points of sides AB and BC of a parallelogram ABCD.DX and DY intersect AC at M and N respectively .If AC = 4.5 ,find MN. 2) The angle between two altitudes of a parallelogram through the same vertex of an obtuse angle of the parallelogram is 60 degree .Find the angles of the parallelogram. plzzz answer this question sir. ABCD is parallelogram. X and Y are the mid point of the sides AB and BC respectively. DX and DY intersect AC in M and N respectively.

In ΔABC, X and Y are the mid point of AB and BC respectively. In ΔAOB,

X is the mid point of AB and XS || OM.  (XY || AC)

∴ S is the mid point of OB.  (Converse of mid point theorem)

⇒ OS = SB

We know that, diagonals of the parallelogram bisect each other.

∴ OD = OB

⇒ OD = OS + SB

⇒ OD = 2OS    (OS = SB) ΔDMO ∼ ΔDXP  (AA Similarity) Adding (2) and (3), we get 2. ABCD is a parallelogram. AX ⊥ BC,  AY ⊥ CD and ∠XAY = 60°.

∴ ∠XAY + ∠AYC + ∠C + ∠AXC = 360°

⇒ 60° + 90° + ∠C + 90° = 360°

⇒ ∠C = 360° – 240° = 120°

∠A = ∠C = 120°    (Opposite angles of parallelogram are equal)

Now, BC || AD  (Opposite sides of the parallelogram are parallel)

∴ ∠C + ∠D = 180°  (If two parallel are intersected by a transversal, then the sum of adjacent interior angles is 180°)

⇒ 120° + ∠D = 180°

⇒ ∠D = 180° – 120° = 60°

∠B = ∠D = 60°    (Opposite angles of parallelogram are equal)

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