Math, asked by tron4556666, 2 months ago

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

Answered by teakook27
13

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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Answered by SanjuMeena
2

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..

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