2018 Q16 AF : FE = 8:5 Area of ABF : BFE = 8:5 (same height triangle, area = base ratio) Area of BFE = 120 (5/8) = 75 Area of AFD = 75 * (8/5)^2 = 192 (area of similar triangle ratio = side ratio ^2) Let the height of the parallelogram be y In ABD, 8k (y) (1/2) = 192 + 120 = 312 In ABE 5k (y) (1/2) = 195 CDEF = 8ky - 312 - 195 + 120 = 312*2 - 312 - 195 + 120 = 237
So long as the order of vertices is ABCD, you can construct the parallelogram in any way you like and you still get the right conclusion despite awkward shapes. The steps: (1) let equal angles be x (2) angle BDE = x (alt angles // lines) (3) triangle BDE isosceles (base angles equal to x) (4) BE = DE (equal sides of isosceles triangle BDE) (5) triangle ABE isoceles triangle (equal sides BE = DE = AE as given) (6) angle BAE = x (base angles of isosceles triangle equal) (7) triangle ADB isosceles triangle (base angles equal to x) (8) AB = BD (equal sides of isoceles triangle ADB) (hence I correct) (9) angle sum of triangle ADB = 4x = 180 and given angle ABC = 3x = 135) (hence II correct) (10) triangle ABE and triangle DBE are congruent (SAS: AB = BD as equal sides of isosceles triangle ADB, common side BE, inclusive angles equal to x as given) (hence III correct). The whole derivation and conclusion are straight forward and independent of the shape drawn. Don't waste time searching for a correct shape. The actual reason of high failure rate is students rely too much on imprecise picture of shape instead of geometric rules that are always true. Learn the rules then use them.
The explanation for the last step of 3rd problem i.e. A1:A2 = 7:5 should not be area ratio of hour-glass similar triangles (A1 & A2 do not form hour-glass.) but area ratio of adjacent triangles on DB where DH:HB = 7:5.
