Question

in the figure ABCD is a parallelogram in which AC is a diagonal G E and F are the midpoints of Ab bc and ac respectively if triangle GEF is an equilateral triangle then prove that paralleogram ABCD is only a rhombus not a square

Answer 1

Answer:

ABCD is ∥gm

AB∥CD

AE∥FC

⇒AB=CD

21AB=21CD

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..