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.
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°.
AXCY is a quadrilateral.
∴ ∠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)