ABCD is a square. E, F, G and H are the midpoints of AB, BC, CD and AD respectively. Prove that EFGH is square.

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GHEF

EFGH

FEHG

HGFE

answer is B.

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Detailed Solution

Before solving the question, we need to understand the definitions of Midpoint Theorem and Reversal of Midpoint Theorem.
Consider triangle ABC as shown in the figure below.
Let D be the midpoint of AB.

The converse of the midpoint says that if E is the midpoint of BC then DE is parallel to AB and

D E = 12 A B

.
The statement on the contrary is
"The segment joining the midpoints of two sides of a triangle is parallel to the third side and is half of it"
Given: square ABCD. E, F, G, and H are the midpoints of AB, BC, CD, and AD, respectively.
For proof: EFGH is a square.
Construction: Connect AC.
Evidence: In triangle ADC, G is the midpoint of CD and H is the midpoint of AD.
We know that in a triangle, the segment joining the midpoints of two sides of the triangle is parallel to the third side and one half of it.
Therefore, by inverting the sentence about the center, GH||AC and

GH= 12 AC

.
Similarly FE||AC and

F E = 12 A C

Hence, we have FE||GH and

F E = G H = 12 A C

Similarly HE||GF and

HE=GF= 12 BD

Since ABCD is a square, we have AC = BD {Because the diagonals of a square are equal}
So we have FE||GH and HE||GF and FE= GH = HE = GF.
EFGH is therefore a rhombus.
Now in the triangle AHE, AH = AE =