36. ABCD is a Parallelogram E and F are the mid-Points of BC and AD respectively. Show that
the segments BF and DE trisect the diagonal AC.
Answers
Answer:
ABCD is ∥gm
AB∥CD
AE∥FC
⇒AB=CD
2
1
AB=
2
1
CD
AE=EC
AECF is ∥gm
In △DQC
F is mid point of DC
FP∥CQ
By converse of mid point theorem P is mid point of DQ
⇒DP=PQ (1)
∴AF and EC bisect BD
In △APB
E is mid point of AB
EQ∥AP
By converse of MPT ( mid point theorem )
Q is mid point of PB
⇒PQ=QB (2)
By (1) and (2)
⇒PQ=QB=DP
AF and EC bisect BD..
solution
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Answer:
Answer
ABCD is ∥gm
AB∥CD
AE∥FC
⇒AB=CD
2
1
AB=
2
1
CD
AE=EC
AECF is ∥gm
In △DQC
F is mid point of DC
FP∥CQ
By converse of mid point theorem P is mid point of DQ
⇒DP=PQ (1)
∴AF and EC bisect BD
In △APB
E is mid point of AB
EQ∥AP
By converse of MPT ( mid point theorem )
Q is mid point of PB
⇒PQ=QB (2)
By (1) and (2)
⇒PQ=QB=DP
AF and EC bisect BD..