Vrijeme: 02:04

Beskorisna geogebra | Useless geogebra #2

Dan je jednakokračan \triangle ABC s osnovicom BC duljine 20 i krakovima duljine 21. Točke D, E su redom na stranicama AB, AC takve da je dužina DE paralelna osnovici. Točka F se nalazi unutar \triangle ABC tako da je četverokut ADFE tetivan i vrijedi DF=3, FE=4. Omjer površina četverokuta ADFE i \triangle ABC se može izraziti kao razlomak \frac{a}{b}, gdje su a i b relativno prosti prirodni brojevi. Koliko je a+b?
An isosceles \triangle ABC with a base BC of length 20 and legs of length 21 is given. The points D, E are respectively on the sides AB, AC such that the DE is parallel to the base. Point F is inside \triangle ABC so that quadrilateral ADFE is cyclic and DF=3, FE=4 holds. The ratio of the areas of quadrilateral ADFE and \triangle ABC can be expressed as a fraction \frac{a}{b}, where a and b are relatively prime positive integers. What is the value of a+b?