Advertisements
Advertisements
प्रश्न
P and Q are the points of trisection of the diagonal BD of a parallelogram AB Prove that CQ is parallel to AP. Prove also that AC bisects PQ.
Advertisements
उत्तर
We know that, diagonals of a parallelogram bisect each other
∴OA = OC and OB = OD
Since P and Q are point of intersection of BD
∴BP = PQ = QD
Now, OB = OD and BP = QD
⇒ OB - BP = OD - QD
⇒ OP = OQ
Thus in quadrilateral APCQ, we have
OA = OC and OP = OQ
⇒ diagonals of quadrilateral APCQ bisect each other
∴APCQ is a parallelogram
Hence AP || CQ
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 .
ABCD is a square. AC and BD intersect at O. State the measure of ∠AOB.
In a parallelogram ABCD, write the sum of angles A and B.
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
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=
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)
