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A triangle is a point that is equidistant from each vertex of the triangle______ The intersection of the two
∵ the points with equal distance to one side of the triangle should be on the vertical bisector on this side, and the points with equal distance to the other side of the triangle should be on the vertical bisector on this side. There is an intersection point between the two vertical bisectors. From the equivalent substitution, we can see that the points with equal distance to each vertex of the triangle are the intersection points of the vertical bisectors on the three sides of the triangle The vertical bisector of three sides
What is the point with equal distance to three given vertices of triangle ABC?
Question: the angle between the vertical bisector of one waist and the straight line of the other waist of an isosceles triangle is known to be 40 degrees. Find the vertex angle of the isosceles triangle,
It is known that ⊙ ABC is an acute triangle, ⊙ o passes through points B and C, and intersects with edges AB and AC at points D and e respectively. If the radius of ⊙ o is equal to the radius of the circumscribed circle of ⊙ ade, then ⊙ o must pass through ()
A. Inner B. outer C. center of gravity D. vertical
As shown in the figure, connecting be. ∵ ABC is an acute triangle, and ∵ BAC and ∵ Abe are all acute angles. The radius of ∵ o is equal to the radius of the circumscribed circle of △ ade, and De is the common chord of the two circles, ∵ BAC = ∵ Abe. ∵ bec = ∵ BAC + ∵ Abe = 2 ∵ BAC. If the outer center of △ ABC is O1, then ∵ bo1c = 2 ∵ BAC, ∵ o must pass the outer center of △ ABC
In the acute triangle ABC, ab = 13, AC = 13, s triangle ABC = 60
Finding the value of TGC
Let BC = 2A; ah = h; ∵ AB = AC = 13; BH = ch = 1 / 2BC = a; ab ^ 2 = ah ^ 2 + BH ^ 2 = H ^ 2 + A ^ 2; H ^ 2 + A ^ 2 = 13 ^ 2 = 169 --- - (1) s △ ABC = 1 / 2BC * ah = 1 / 2 * 2A * H = ah; ah = 60; --- - (2); (1) - 2 * (2) get: H ^ 2 + A ^ 2-2ah = (H-A) ^ 2 = 169-2 * 60 = 49; H-A = 7
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