Дан треугольник abc проведена медиана cm cm=1/2ab определите вид треугольника

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Через середину K медианы BM треугольника ABC и вершину A проведена прямая, пересекающая сторону BC в точке P. Найдите отношение площади треугольника ABK к площади четырёхугольника KPCM.
Решение.
Через вершину В проводится прямая II АС. АР продолжается за точку Р до
пересечения с этой прямой в точке Е.
Итак, ВЕ || AC;
Треугольники ЕВК и АКМ равны по второму признаку (у них углы ВКЕ и АКМ равны как вертикальные, <ЕВK=<KMА как накрест лежащие при параллельных АМ и ВЕ и секущей ВМ, а ВК=КМ - дано), значит ЕВ = АМ.
Отсюда ЕВ = АС/2; (так как ВМ - медиана и АМ=0,5АС).
Треугольники ЕВР и АСР подобны по двум углам (углы ВPE и АКМ равны как
вертикальные, <EAC=<BEA, как накрест лежащие при параллельных АС и ВЕ и
секущей АЕ), поэтому ВР/РС = ЕВ/АС = 1/2 (так как ЕВ = 1/2*АС).
Отсюда РС = 2ВР.  То есть ВС равна ВР+2ВР = 3ВР или ВС разделена точкой Р на части 1/3 и 2/3. Итак, СР = ВС*2/3. Площадь треугольника АСР равна площади треугольника АВС минус площадь треугольника АВР. По известной формуле S=1/2*BC*h имеем площадь тр-ка АВС.
Заметим, что у тр-ков АВС, АВР и АРС высота h, проведенная к основанию ВС (ВР,РС) одна и та же, можем сказать что их площади относятся, как их основания, то есть 1:1/3:2/3. Тогда Sacp = S*2/3; (S - площадь треугольника АВС). Поскольку площадь треугольника АВМ равна половине площади АВС, а площадь АКМ равна половине АВМ (из свойства медианы треугольника,
которая делит тр-к на два равновеликих), то
Sakm = (1/2)Sabm = (1/2)*(1/2)*Sabс = (1/4)Sabс.
Площадь четырехугольника КРСМ равна площади треугольника ACP минус площадь треугольника AKM. Подставляем известные нам величины и получим:
Skpсm=(2/3)Sabc-(1/4)Sabc = (5/12)Sabc.
Отношение Sabk/Skpcm = (1/4):(5/12) = 3/5. (Sabk=Sakm=(1/4)Sabс по свойству
медианы АК тр-ка АВМ, которая делит тр-к на два равновеликих).
ответ: Sabk/Skpcm=3/5.