Advertisements
Advertisements
Question
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
Solution

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
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 a parallelogram ABCD, if `∠`B = 135°, determine the measures of its other angles .
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.
PQRS is a quadrilateral, PR and QS intersect each other at O. In which of the following case, PQRS is a parallelogram?
∠P = 100°, ∠Q = 80°, ∠R = 95°
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, 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)
Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle.
