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In the quadrilateral ABCD, the diagonal AC bisects ∠ bad, AB > ad
Find a point E on AB so that AE equals ad, and then connect De
Ad = AE, AC = AC, angle EAC = angle CAD, so triangle AEC is equal to triangle ADC
So DC = EC; EB = AB ad
As shown in the figure, e is a point on the extension line of the edge ab of the parallelogram ABCD, and de intersects BC with F. if s △ ABF = 3, calculate the area of △ efc
CD is parallel to AE, and the angle BFE is equal to the angle CFD, so the triangle bef is similar to the triangle CDF. CF / BF = CD / be because of the same base and height, the area ratio of triangle ABF to triangle bef = AB / be, the area ratio of triangle CEF to triangle bef = CF / BF = CD / be and ab = CD, so the areas of triangle ABF and triangle CEF are equal three times
As shown in the figure, e is a point on the extension line of the edge ab of the parallelogram ABCD, and de intersects BC with F. it is proved that the area of the triangle ABF is equal to the area of the triangle efc
Through E and a, make vertical lines en, am intersecting BF at n, m by DC / / EB, we can know FC: BF = FD: Fe by AD / / BF, EF: DF = EB: ab by AM / / BN, we can know AB: EB = am: en, so FC: BF = am: en, the area of triangle ABF is equal to 1 / 2am * BF, the area of triangle EFC is equal to 1 / 2EN * CF, so the area of triangle ABF is equal to the area of triangle efc
RELATED INFORMATIONS
- 1. It is known that e is the edge of the parallelogram ABCD, AB extends to the upper point, de intersects BC with F. proof: the area of the triangle ABF is equal to the area of the triangle efc Mathematics in grade two
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