Question

In a parallelogram ABCD, E and F are the midpoints of sides AB and CD respectively. Show that the line segment AF And EC trident the diagonal BD

Answer 1

Given ABCD is a parallelogram
Hence AB || CD
⇒ AE || FC
Also AB = CD (Opposite sides of parallelogram ABCD)
⇒ AE = FC (Since E and F are midpoints of AB and CD)
In quadrilateral AECF, one pair of opposite sides are equal and parallel.
∴ AECF is a parallelogram.
⇒ AF || EC (Since opposite sides of a parallelogram are parallel)
In ΔDPC, F is the midpoint of DC and FQ || CP
Hence Q is the midpoint of DQ by converse of midpoint theorem.
⇒ DQ = PQ → (1)
Similarly, in ΔAQB, E is the midpoint of AB and EP || AQ
Hence P is the midpoint of DQ by converse of midpoint theorem.
⇒ BP = PQ → (2)
From equations (1) and (2), we get
BP = PQ = DQ
Hence, the line segments AF and EC trisect the diagonal BD of parallelogram ABCD



hope it helpz.

Answer 2

FC =1/2 DC
EA=1/2 AB
FC=AE
FC is parallel to AE (as AB is parallel CD)
FCEA is a parallelogram
in triangle DCQ = F Is the mid point and FP is parallel to QC
therefore P is the mid point of DQ
In triangle ABP E is the mid point of AB , AP is parallel to EQ
Therefore Q is also a mid point
Therefore DP=PQ=QB.
(P and Q are points where the line AF and CE intersect DB)