Advertisements
Advertisements
Question
The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.
Solution
Since ABCD is a parallelogram
∴AB || DC and AB = DC
⇒EB || DF and `1/2` AB =`1/2` DC
⇒EB || DF and EB = DF
EBFD is a parallelogram
APPEARS IN
RELATED QUESTIONS
In Fig., below, ABCD is a parallelogram in which ∠A = 60°. If the bisectors of ∠A and ∠B meet at P, prove that AD = DP, PC = BC and DC = 2AD.
In Fig. below, AB = AC and CP || BA and AP is the bisector of exterior ∠CAD of ΔABC.
Prove that (i) ∠PAC = ∠BCA (ii) ABCP is a parallelogram
In a parallelogram ABCD, if ∠D = 115°, then write the measure of ∠A.
We get a rhombus by joining the mid-points of the sides of a
The figure formed by joining the mid-points of the adjacent sides of a rhombus is a
The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a
In a quadrilateral ABCD, ∠A + ∠C is 2 times ∠B + ∠D. If ∠A = 140° and ∠D = 60°, then ∠B=
In the given figure, ∠A = 64°, ∠ABC = 58°. If BO and CO are the bisectors of ∠ABC and ∠ACB respectively of ΔABC, find x° and y°
In the given Figure, if AB = 2, BC = 6, AE = 6, BF = 8, CE = 7, and CF = 7, compute the ratio of the area of quadrilateral ABDE to the area of ΔCDF. (Use congruent property of triangles)
ABCD is a square, diagonals AC and BD meet at O. The number of pairs of congruent triangles with vertex O are
