(см. объяснение)
Пошаговое объяснение:
Покажем сначала, что биссектрисы AM и BM пересекаются под прямым углом. Действительно, пусть ∠ABC=α и ∠BAD=β. Тогда α+β=180°. Так как биссектриса делит угол пополам, то верно, что ∠ABM+∠BAM=α/2+β/2=90°, поэтому и ∠BMA=90°.
Опустим из точки M перпендикуляр на сторону BC. Получим ME⊥BC. Тогда ΔBMA~ΔBEM по двум углам. Из подобия треугольников следует, что AB/BM=AM/ME.
Опустим из точки M перпендикуляр на сторону AD. Получим, MT⊥AD. Тогда ΔBMA~ΔATM по двум углам. Из подобия треугольников следует, что AB/AM=BM/MT, то есть AB/BM=AM/MT.
Так как AB/BM=AM/ME и AB/BM=AM/MT, то верно, что AM/ME=AM/MT или ME=MT.
Так как расстояния от точки M до прямых BC и AD одинаковы, то точка M лежит на средней линии трапеции.
Применив аналогичное рассуждение, получаем, что точка N тоже лежит на средней линии трапеции.
Тогда MN - это часть средней линии трапеции, то есть MN||BC и MN||AD.
Проведем среднюю линию трапеции FG. По определению FG=(7+12)/2=19/2.
Так как треугольники ΔBMA и ΔCND прямоугольные, а F и G - середины их гипотенуз AB и CD, то FM и GN - это медианы, равные AB/2 и CD/2 соответственно, то есть FM=5/2 и GN=4.
Понятно, что MN=FG-FM-GN, а значит MN=19/2-5/2-4=3.
Задача решена!
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Пошаговое объяснение:
а+в=36-15
а²+в²=15²
а=21-в
(21-в)+в²=225
441-42в+в²+в²-225=0
2в²-42в+216=0
в²-21в+108=0
в=21+√(441-432) = 21+3=12см один катет
2 2
21-12=9см второй