in a parallelogram ABCD, E and F are the midpoints of sides AB and CD respectively . AF and EC meet the diagonal BD of length 12cm at P and Q the length of PQ is
Answers
Step-by-step explanation:
Given: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively
To prove: the line segments AF and EC trisect the diagonal BD.
Proof : In quadrilatweral ABCD,
AB=CD (Given)
\frac{1}{2}AB=\frac{1}{2}CD
\Rightarrow AE=CF (E and F are midpoints of AB and CD)
In quadrilateral AECF,
AE=CF (Given)
AE || CF (Opposite sides of a parallelogram)
Hence, AECF is a parallelogram.
In \triangle DCQ,
F is the midpoint of DC. (given )
FP || CQ (AECF is a parallelogram)
By converse of midpoint theorem,
P is the mid point of DQ.
DP= PQ....................1
Similarly,
In \triangle ABP,
E is the midpoint of AB. (given )
EQ || AP (AECF is a parallelogram)
By converse of midpoint theorem,
Q is the midpoint of PB.
OQ= QB....................2
From 1 and 2, we have
DP = PQ = QB.
Hence, the line segments AF and EC trisect the diagonal BD.
Answer:
parallogram ABCD.
BC is parallel
Step-by-step explanation:
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