As shown in the figure, △ ABC rotates counterclockwise to the position of △ ab ′ C ′ so that AC bisects BB ′

It is proved that in △ ABB ', ab = ab', AC bisects BB ', AC is the vertex bisector of isosceles △ ABB', i.e., ∠ BAC = ∠ B ′ AC, in △ AMC and △ AMC ', AC = AC', MAC = ∠ Mac ', am = am, ｜ △ AMC ≌ AMC ′, MC = MC', so ab ′ bisects CC '

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1. Ask math question AA + BB + CC = ABC What are the ABC numbers?
2. Given a + B + C = 0, ABC = - 15, find AA (B + C) + BB (a + C) + CC (a + b)
3. As shown in the figure, in the isosceles right angle △ ABC, ∠ ABC = 90 °, point D is on AC, and △ CBE is obtained after △ abd is rotated 90 ° clockwise around vertex B. (1) calculate the degree of ∠ DCE; (2) when AB = 4, ad: DC = 1:3, calculate the length of de
4. If the right angle vertex C of RT △ ABC is in plane a, the hypotenuse ab ‖ a, and the distance between AB and a is root 6, the projective points of a and B in a are A1 and B1 respectively, and A1C = 3, B1C = 4, then ∠ a1cb1=
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6. Ad is the bisector of triangle ABC, De is parallel to AC, de intersects AB with E, DF is parallel to AB, DF intersects AC with F. what is the relationship between angle EDA and angle ADF?
7. In a right triangle ABC, angle a is a right angle, point D is the midpoint of BC, e is on AB, f is on AC, De is perpendicular to DF, and we prove that EF square = be square + CF square
8. In △ ABC, ab = AC, D is the midpoint of BC, e and F are on AB and AC respectively, and △ DEF is an equilateral triangle
9. As shown in the figure, △ ABC is a right triangle, ∠ ACB = 90, ad is the bisector of ∠ BAC, de ⊥ AB, BC = 8, CD = 3, the length of 1.be, 2. The area of △ ABC
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17. The area of △ AEF is equal to the area of trapezoidal EFBC
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