Question

solve it fast please​

Answer 1

$\blue\bigstar$Given:

• ABCD is a parallelogram where E and F are the mid - points of sides AB and CD respectively.

$\pink\bigstar$To prove:

• AF and EC trisect BD i.e. BG = QP = DP

$\red\bigstar$ Proof:

ABCD is a parallelogram,

$\therefore$ AB || CD

(Opposite sides of a Parallelogram are equal)

$\implies$ AE || CF ( Parts of Parallel lines are parallel )

and AB = CD

( Opposite sides of a Parallelogram are equal )

$\sf\implies \dfrac{1}{2} \: AB \: = \: \dfrac{1}{2} \: CD \:\: i.e. \: \:AE \: = \: CF$

$\therefore$AE = CF

( Given F is the mid - point of CD and E is the mid point of AB )

$\implies$AECF is a Parallelogram.

In $\triangle$ABP,

$\because$ E is the mid point of AB and EQ || AP,

$\implies$ BQ = QP .......(i)

(Converse of Mid - point theorem )

In $\triangle$DQC,

$\because$F is the mid point of DC and FQ || CQ

$\implies$DP = PQ .........(ii)

( Converse of Mid - point Theorem )

From (i) and (ii),

$\boxed{\sf BQ \: = \: PQ =\: DP\:}$

$\implies$BD is triesected at P and Q.

Hence, proved.