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As shown in the figure, in parallelogram ABCD, E is a point on AD. Lines CE and BD intersect at point F, and the areas of triangles BCF and DCF are 6 and 3 respectively； What is the area of the red part?
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The answer is 7.5. Draw two auxiliary lines to connect AF, and draw a perpendicular line GH through F to form a parallelogram. BCF=6, DCF=3; thus BCD=6+3=9. The perpendicular line FH equals GH*2/3, and this is key ； Starting the calculations: ADF = 9*1/3 = 3, ABF = 9 – 3 = 6. Meanwhile, DEF = CDE*1/3 = (DEF + DCF)*1/3; 2*DEF = DCF = 3, so DEF = 1.5. AEF = ADF – DEF = 3 – 1.5 = 1.5 ； ABEF=ABF+AEF=6+1.5=7.5.
There’s no need for such complexity; it’s sufficient to note that △BCF and △DFE are similar, and that DF/BF = 1/2.
It equals 3√2 + 3. Since the triangles are similar, once the area is known, it’s easy to determine the ratios of the sides and heights. The area of the shaded region is then equal to the area of the larger triangle minus 3
Since △BCF and △DFE are similar, it can be determined that the area of triangle DEF is 1.5. Therefore, the area of the shaded region ABFE equals the area of triangle ABD minus the area of triangle DEF, which is 9 – 1.5 = 7.5