There is a point P in the equilateral △ ABC. The distance from it to the three vertices a, B and C is 1, root 2 and root 3 respectively. Find the degree of ∠ ABC

There is a point P in the equilateral △ ABC. The distance from it to the three vertices a, B and C is 1, root 2 and root 3 respectively. Find the degree of ∠ ABC

The triangle ABP is rotated 60 degrees around point a in reverse time, AB and AC coincide, and point P falls outside the triangle, that is, H,
The results show that △ ABP is congruent △ ACh, ∠ APB = ∠ AHC, AP = ah, BP = ch = 2 under the root, and ∠ PAH = 60 degrees, so △ APH is an equilateral triangle, ∠ AHP = 60 degrees, pH = AP = 1, in △ CPH, CP = 3 under the root, ch = 2 under the root, pH = 1, so CP ^ 2 = ch ^ 2 + pH ^ 2,
So △ CPH is a right triangle, ∠ CHP = 90, ∠ APB = ∠ AHC = ∠ AHP + ∠ CPH = 60 + 90 = 150 degrees
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2. Given that the distance from the vertex of acute angle △ to the perpendicular is equal to the radius of its circumscribed circle, then the degree of ∠ is

It is 60 degrees. The distance from the perpendicular to the three sides is equal, and the distance to the vertex is also equal

The point with equal distance to the three vertices of △ ABC is the intersection of △ abc()
A. Trilateral center line B. trilateral bisector C. trilateral upper height D. trilateral vertical bisector

The point with equal distance to the three vertices of the triangle is the intersection point of the vertical bisector of the three sides of the triangle, so D