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№1.
1) АВ=АD
2) BC=CD
3) AC - совместная сторона ΔABC и ΔADC.
Значит, ΔАВС=ΔАDC по 3 признаку равенства треугольников.
Поэтому угол ВАС = углу СAD = угла 1/2 ВАD = 80° : 2 = 40°
№2.
1) угол 1 = углу 2
2) угол 3 = углу 4
3) ВD - совместная сторона ΔABD и ΔCDB
Отсюда ΔABD=ΔCDB по 2 признаку равенства треугольников. Значит, АВ = CD = 12 см.
ответ: 12 см.
№3.
1) АО=ВО
2)СО=OD
3) угол СОА = углу BOD (вертикальные)
Тогда ΔAOC=ΔBOD по 1 признаку равенства треугольников. Поэтому угол САО=углу DBO= 60° и BD=AC=14 дм.
ответ: угол DBO=60°; АС=14дм.
Следовательно если эта медиана делит сторону на 2 равные части, то она действительно является медианой)