Question

exercise #3: in the diagram at the right, square abcd is drawn and the midpoints of its sides have been marked and connected with segments to form quadrilateral efgh. give an explanation for why efgh must also be a square.

Explanation:

Step1: Prove congruent right - angled triangles

Since \(ABCD\) is a square, \(AB = BC=CD = DA\), and \(\angle A=\angle B=\angle C=\angle D = 90^{\circ}\). Let \(AB = BC=CD = DA = 2x\). Then \(AE=EB=BF=FC=CG=GD=DH=HA=x\).
By the Pythagorean theorem, in right - triangle \(AEH\), \(EH=\sqrt{AE^{2}+AH^{2}}=\sqrt{x^{2}+x^{2}}=\sqrt{2}x\). Similarly, in right - triangle \(BEF\), \(EF = \sqrt{BE^{2}+BF^{2}}=\sqrt{x^{2}+x^{2}}=\sqrt{2}x\); in right - triangle \(CFG\), \(FG=\sqrt{FC^{2}+CG^{2}}=\sqrt{x^{2}+x^{2}}=\sqrt{2}x\); in right - triangle \(DGH\), \(GH=\sqrt{GD^{2}+DH^{2}}=\sqrt{x^{2}+x^{2}}=\sqrt{2}x\). So \(EH = EF=FG = GH\).

Step2: Prove right - angles in \(EFGH\)

\(\angle AEH+\angle BEF = 90^{\circ}\). Because \(\triangle AEH\cong\triangle BEF\) (by \(SAS\): \(AE = BE\), \(\angle A=\angle B\), \(AH = BF\)), \(\angle AEH=\angle BEF\). Let \(\angle AEH=\alpha\) and \(\angle BEF=\alpha\), then \(\angle HEF=180^{\circ}- 2\alpha\). Since \(\angle A = 90^{\circ}\) and \(\triangle AEH\) is a right - isosceles triangle (\(AE = AH\)), \(\alpha = 45^{\circ}\), so \(\angle HEF = 90^{\circ}\). Similarly, \(\angle EFG=\angle FGH=\angle GHE = 90^{\circ}\)

Answer:

Since all the sides of quadrilateral \(EFGH\) (\(EH, EF, FG, GH\)) are equal (\(EH = EF=FG = GH=\sqrt{2}x\)) and all its interior angles (\(\angle HEF,\angle EFG,\angle FGH,\angle GHE\)) are right - angles (\(90^{\circ}\)), by the definition of a square (a quadrilateral with four equal sides and four right - angles), \(EFGH\) is a square.