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In ABC, AB is equal to AC, BF is the height on the side of AC, and the angle FBC is equal to 1 / 2 angle BAC
Let C = angle ABC = X
Because BF is high
So the angle BFC is 90 degrees
So the angle FBC is 90 ° - X
Because: angle ABC + angle c + angle BAC = 180 degree
So: angle BAC = 180 ° - 2x = 2 (90 ° - x)
So: the angle FBC is equal to 1 / 2 of the angle BAC
As shown in figure Z-15, in △ ABC, ∠ FBC = ∠ ECB = 1 / 2 ∠ a means be = CF
Please, there are no pictures
As shown in the figure, the isosceles right angle △ ABC, Ao is the middle line on the hypotenuse, D is the point on AC, OE ⊥ od intersects AB with E. please explain the reason why od = OE
It is proved that: ∵ △ ABC is isosceles right triangle, ∵ C = ∠ B = 45 °. ∵ Ao is the middle line on the hypotenuse, ∵ Ao = co = Bo = 12bc, ∵ Cao = ∠ Bao = 45 °, ∵ AOC = 90 °, ∵ C = ∠ EAO. ∵ OE ⊥ OD, ∵ EOD = ∠ EOA + ∠ DOA = 90 °. ? cod + ∠ AOD = 90 ° and ∵ cod = ∠ AOE
As shown in the figure, the line segments BD and CE in △ ABC intersect at the point O. ob = od and OC = 2oe are known. Let the areas of △ BOE, △ BOC, △ COD and quadrilateral aeod be S1, S2, S3 and S4 respectively. (1) find the value of S1: S3; & nbsp; & nbsp; & nbsp; (2) if S2 = 2, find the value of S4
(1) The edge ob of ∵ BOC and the edge od of ∵ doc have the same height. Let H be the height, ∵ s2s3 = 12 × ob × H12 × OD × H = obod, ∵ ob = OD, ∵ S2 = S3, the edge OE of ∵ BOE and the edge OC of ∵ BOC have the same height. Let a be the height, ∵ S1S2 = 12 × OE × A12 × OC × a = oeoc, ∵ OC = 2oe, ∵ S2 = 2S1, ∵ S3 = 2S1, ∵ S1: S3 = 1:2. (2) connect OA, ? S2 = 2, ∵ S1 = 1, S3 = 2, let ? AOD have the same height The area is x, ∵ ob = OD, ∵ Bao is x, ∵ AOE is X-1, ∵ OC = 2oe, ∵ s △ AOC = 2S △ AOE, ∵ x + 2 = 2 (x-1). The solution is x = 4, ∵ S4 = 4 + 4-1 = 7
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