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Question 7 (This is the square of mathematics learning method of Hunan Education Press.
The perimeter of the quadrilateral EFGH can be found.

It is proved that the intersection of EF, GH and BD is p and q respectively.

∫EH//BD//FG

∴ △AEH, △ABD, △CFG and △CBD are isosceles right triangles.

△AEH?△CFG

AE=CF

AB = BC

∴? △BEF is an isosceles right triangle.

EPQH is rectangular.

EH=PQ？ EP=BP？ Headquarters =DQ

EP+EH+HQ=BD

In the same way; In a similar way

FP+FG+GQ=BD

Circumference of quadrilateral EFGH = 2BD

According to Pythagorean theorem

BD=√2AB=√2* 1/4 square ABCD perimeter = √2a.

Circumference of quadrilateral EFGH = 2√2a

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