Advertisements
Advertisements
Question
Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ (Figure). Show that AC and PQ bisect each other.
Advertisements
Solution
Given: ABCD is a parallelogram and AP = CQ
To show: AC and PQ bisect each other.
Proof: In ΔAMP and ΔCMQ,
∠MAP = ∠MCQ ...[Alternate interior angles]
AP = CQ ...[Given]
And ∠APM = ∠CQM ...[Alternate interior angles]
∴ ΔAMP ≅ ΔCMQ ...[By ASA congruence rule]
⇒ AM = CM ...[By CPCT rule]
And PM = MQ ...[By CPCT rule]
Hence, AC and PQ bisect each other.
Hence proved.
APPEARS IN
RELATED QUESTIONS
The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O. If ∠DAC = 32º and ∠AOB = 70º, then ∠DBC is equal to ______.
Diagonals of a quadrilateral ABCD bisect each other. If ∠A = 35º, determine ∠B.
E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. Show that BFDE is a parallelogram.
In a parallelogram ABCD, AB = 10 cm and AD = 6 cm. The bisector of ∠A meets DC in E. AE and BC produced meet at F. Find the length of CF.
A diagonal of a parallelogram bisects one of its angles. Show that it is a rhombus.
In the following figure, AB || DE, AB = DE, AC || DF and AC = DF. Prove that BC || EF and BC = EF.
If diagonals of a quadrilateral bisect each other, it must be a parallelogram.
The point of intersection of diagonals of a quadrilateral divides one diagonal in the ratio 1:2. Can it be a parallelogram? Why or why not?
Two sticks each of length 7 cm are crossing each other such that they bisect each other at right angles. What shape is formed by joining their end points? Give reason.
