Advertisements
Advertisements
प्रश्न
The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.
Advertisements
उत्तर

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
संबंधित प्रश्न
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 a parallelogram ABCD, if `∠`B = 135°, determine the measures of its other angles .
In a ΔABC median AD is produced to X such that AD = DX. Prove that ABXC is a
parallelogram.
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, the bisector of ∠A also bisects BC at X. Find AB : AD.
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
ABCD is a parallelogram in which diagonal AC bisects ∠BAD. If ∠BAC = 35°, then ∠ABC =
In a quadrilateral ABCD, ∠A + ∠C is 2 times ∠B + ∠D. If ∠A = 140° and ∠D = 60°, then ∠B=
ABCD is a square, diagonals AC and BD meet at O. The number of pairs of congruent triangles with vertex O are
