Докажем, что из условия FC+CH=EA+AG следует, что △EBF - равнобедренный.
Рассмотрим равные треугольники E1BF1 и HDG (E1F1||GH). Условие очевидно выполняется (F1C=AG, E1A=CH).
Теперь рассмотрим данный отрезок EF. Если условие выполняется, то EE1=FF1. Тогда EE1F1F - равнобедренная трапеция, углы при основании равны. Следовательно △EBF также равнобедренный.
BE=BF=2x, AE=3x, CF=9x
BF=2x=4 => x=2
FC+CH=EA+AG => 9x+3 =3x+AG => AG =18+3-6 =15
In ABCD parallelogram E, F, G, and H are the midpoints of AB, BC, CD, and DA respectively. Area of ABCD is 20 sq. unit. What's the area of PQRS?
(Correcting my earlier answer)
Let's assume that ABCD is a square (which is certainly a parallelogram). Then PQRS is also a square surrounded by 4 congruent right triangles (ABQ, BCR, CDS, DAP).
Area of square PQRS = 20 - 4 * (Area of ABQ)
AB = BC = CD = DA = sqrt(20)
ABQ is similar to AFB, so AQ = 2 * BQ
The hypotenuse of ABQ is sqrt(20) and the legs are BQ and 2 * BQ, so by the Pythagorean Theorem: BQ^2 + (2 * BQ)^2 = 20, or:
5 * BQ^2 = 20 => BQ = sqrt(4) = 2
Area of ABQ = 0.5 * base * height = 0.5 * AQ * BQ = 0.5 * 4 * 2 = 4
Area of square PQRS = 20 - 4 * (Area of ABQ) = 20 - 4 * 4 = 4
So, assuming that the answer is the same for all parallelograms (square or not), it's 4.
I don't have the proof for that.