Question

answer = 50 points rewards
spammer = report
solve 4 , 5 and 6​

Answer 1

4)

ABCD is ∥gm

AB∥CD

AE∥FC

⇒AB=CD

AB= 1/2

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..

5)

In △ADC,S is the mid-point of AD and R is the mid-point of CD

In △ABC,P is the mid-point of AB and Q is the mid-point of BC

Line segments joining the mid-points of two sides of a triangle is parallel to the third side and is half of of it.

∴SR∥AC and SR= 1/2

AC ....(1)

∴PQ∥AC and PQ= 1/2

AC ....(2)

From (1) and (2)

⇒PQ=SR and PQ∥SR

So,In PQRS,

one pair of opposite sides is parallel and equal.

Hence, PQRS is a parallelogram.

PR and SQ are diagonals of parallelogram PQRS

So,OP=OR and OQ=OS since diagonals of a parallelogram bisect each other.

Hence proved.

Answer 2

Answer:

(4)

ABCDllgm

ABllCD

AEllFC

→AB=CD

$ab = \frac{1}{2}$

CD

AE=EC

AECF is llgm

in∆DQC

F is midpoint of DC

FPllCQ

By Converse of mid point theorem

p is the mid point of DQ

DP=PQ. (1)

AF and EC bisect BD