The angle bisectors of a triangle intersect at a point that is equidistant from the sides of thetriangle
Let’s prove that the angle bisectors of a triangle intersect at a point that is equidistant from the sides of the triangle
Let’s prove that the angle bisectors of a triangle intersect at a point that is equidistant from the sides of the triangle.
Consider a triangle ABC, where the angle bisectors of each angle intersect at a point O.
To prove that point O is equidistant from the sides of the triangle, we need to show that the distances from O to each side of the triangle are equal.
Let AD, BE, and CF be the angle bisectors of angles A, B, and C, respectively. Note that D, E, and F are points on sides BC, AC, and AB, respectively.
We will use the Angle Bisector Theorem to calculate the distances from O to each side of the triangle.
According to the Angle Bisector Theorem, we have:
BD/DC = AB/AC (1)
CE/EA = BC/BA (2)
AF/FB = AC/BC (3)
Now, let’s use the above formulas to calculate the ratios:
BD/DC = AB/AC (from equation 1)
=> BD = (AB * DC) / AC
CE/EA = BC/BA (from equation 2)
=> CE = (BC * AE) / BA
AF/FB = AC/BC (from equation 3)
=> AF = (AC * FB) / BC
Now, let’s calculate the distances from O to each side of the triangle:
Distance from O to side BC = Distance from O to D = BD + CD
= (AB * DC) / AC + (BC * AE) / BA [since BD = (AB * DC) / AC and CE = (BC * AE) / BA]
= (AB * DC + BC * AE) / AC [combining the fractions with the same denominator]
Similarly,
Distance from O to side AC = Distance from O to E = CE + EA
= (BC * AE) / BA + (AC * FB) / BC
= (AC * FB + BC * AE) / BA
And,
Distance from O to side AB = Distance from O to F = AF + FB
= (AC * FB) / BC + (AB * DC) / AC
= (AC * FB + AB * DC) / BC
Since (AB * DC + BC * AE) / AC = (AC * FB + BC * AE) / BA = (AC * FB + AB * DC) / BC,
we can conclude that the distances from O to each side of the triangle are equal.
Therefore, point O, which is the intersection of the angle bisectors of the triangle, is equidistant from the sides of the triangle.
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