In the parallelogram ABCD, the diagonals AC and BD intersect at point K. Prove that the area of the parallelogram ABCD is four times the area of the triangle AKD
Consider the triangles ABC and ACD.
The AC side is common for these triangles, AB = CD and BC = AD (by the property of the parallelogram), therefore, the considered triangles are equal (by the third sign). So their areas are equal, and these areas are equal to half the area of the parallelogram.
Consider the triangle ACD, as we just found out, the area of this triangle is equal to half the area of the parallelogram. The segment DK – is the median (according to the third property of the parallelogram), and accordingly divides this triangle into two equal triangles, i.e. equal in area (median property).
Therefore, the area AKD is equal to half the area of the triangle ACD. SAKD = SACD / 2 = SABCD / 4
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