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It is known that the side length of the equilateral triangle ABC is 4, and point D is a moving point (not coincident with points B and C) on the side BC. It connects AD and makes the vertical bisector of AD and intersects with the sides AB and AC at points E and f respectively, Find the sum of the perimeter of triangle BDE and triangle DCF; Let the length of BD be x, and use the algebraic expression of X to express the perimeter of BDE and CDF;
Because EF is the vertical bisector of AD, the perimeter of triangle BDE and triangle DCF is: be + ed + BD + DF + DC + cf. because AE = ed, AF = DF, the sum of the perimeter of the two triangles can be converted into: be + AE + BD + AF + DC + CF, because AE + EB = AB = 6, AF + FC = AC = 6, BD + DC = BC = 6
It is known that f (x) = the quadratic power of X + 2x + 1, and f (2) = a + B
Finding the value of a + B
The value of X can be obtained when a = 3A + 2B / 2b-3a and 3x = 4 = 6 are specified
There is some relationship between the former and the latter
One more thing, please. I'll get more points
If ▽ / △ = 8 and △ = 3
▽*▲=120 ▲*▼=40
So how big is @?
F (2) = 2 & sup2; + 2 * 2 + 1 = 9, so a + B = 9, a? B = 3A + 2B / 2b-3a, so 3x? 4 = (3 * 3x + 2 * 4) / (2 * 4-3 * 3x) = (9x + 8) / (8-9x) = 69x + 8 = 48-54x, 63x = 40, x = 40 / 63
As shown in the figure, a, O and B are on a straight line, ∠ AOC = 12 ∠ BOC + 30 °, OE bisects ∠ BOC, and calculates the degree of ∠ BOE
∵ OE is the bisector of ∠ BOC, ∵ BOE = ∠ COE = x, ∵ AOC = 180 ° - 2x, according to the meaning of the title: 180 ° - 2x = x + 30 °, the solution: x = 50 °, then ∠ BOE = 50 °
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