#6warmup

1. (St. Petersburg Math Olympiad-2003, City Round, 8.2)
On the hypotenuse AC of a right triangle ABC, a point D is chosen such that BC = CD. On the side BC, a point E is taken such that DE = CE. Prove that AD + BE = DE.
2. (St. Petersburg Math Olympiad-2003, Selection Round, 10.2)
On diagonal AC of a circumscribed quadrilateral ABCD, a point L is taken such that AB = AL. On the ray DC, a point F is chosen such that DB = DF. Point E is symmetric to B with respect to AD. Prove that points F, L, and E lie on a line.
3. (St. Petersburg Math Olympiad-2002, Selection Round, 11.6)
An inscribed circle in triangle ABC touches sides BC, CA, and AB at points A₁, B₁, and C₁, respectively. A line is drawn through point A₁, perpendicular to segment AA₁. This line intersects line B₁C₁ at point X. Prove that line BC bisects segment AX.

[@wicwic3600]

1-ю задачу имхо практичнее (с точки зрения расхода времени на олимпиаде) решать в лоб через счет длин отрезков.
Обозначим гипотенузу AC как d, а угол С как z и просто посчитаем эти отрезки:
AD = AC - DC = d - d*cos(z)
EC = DC/2/cos(z) = (d*cos(z)/2) / cos(z) = d/2
BE = DC - ED = d*cos(z) - d/2
Из чего видно, что необходимое равенство выполняется.

[@OlympiadGeometry]

это недоступное решение для 8 класса в начале года, а так, конечно...