## In square ABCD, the centre O and the midpoint M of side AB are connected with vertices C and D. Prove that areas of △AMD, △BMC, △CDO and a

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## Answers ( No )

The area of a triangle is given by the formula …

… A = (1/2)bh

where b is the base length and h is the height perpendicular to the base.

If we let the length of the sides of the square be represented by s, then we have …

… AM = MB = MO = s/2

and the areas of the triangles are …

… AAMD = (1/2)(s)(s/2) = s²/4

… ABMC = (1/2)(s)(s/2) = s²/4

… ACDO = (1/2)(s)(s/2) = s²/4

and the area of the quadrilateral is the sum of the areas of triangles MOC and MOD, so is …

… AMCOD = (1/2)(s/2)(s/2) +(1/2)(s/2)(s/2) = s²/8 +s²/8 = s²/4

Hence the described areas are equal to each other. (All are s²/4.)