# The perpendicular bisector \overline{AB} in the right triangle \triangle ABC

The perpendicular bisector $$\displaystyle\overline{{{A}{B}}}$$ in the right triangle $$\displaystyle\triangle{A}{B}{C}$$ forms the triangle with the area 3 times smaller than the area of $$\displaystyle\triangle{A}{B}{C}$$. Find the measures of acute angles in $$\displaystyle\triangle{A}{B}{C}$$

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Step 1
The Sides of AB and AC have unit length
Hence, $$\displaystyle\triangle{A}{B}{C}$$ is isosceles with Congruent angles $$\displaystyle\angle{B}$$ and $$\displaystyle\angle{C}$$ NSNK Since $$\displaystyle{m}{\left(\angle{A}\right)}$$ is given as 60 this means that,
$$\displaystyle{m}{\left(\angle{B}\right)}+{m}{\left(\angle{C}\right)}={120}$$
and so, $$\displaystyle{m}{\left(\angle{B}\right)}+{m}{\left(\angle{C}\right)}={60}$$
Therefore $$\displaystyle\triangle{A}{B}{C}$$ is equilateral
Equilateral triangles have three lines of reflective symmetry in which the lines joining each vertex to the midpoint of the opposite side.
This means that CD is a line of symmetry for $$\displaystyle\triangle{A}{B}{C}$$ and so CD is perpendicular to AB.
Step 2
By applying Pythagorean Theorem to the right triangle $$\displaystyle\triangle{A}{D}{C}$$
$$\displaystyle{\left|{A}{D}\right|}{2}+{\left|{C}{D}\right|}{2}={\left|{A}{C}\right|}{2}$$
We Know that D is the midpoint of AB.
B is on the unit circle,
Hence $$\displaystyle{\left|{A}{B}\right|}={1}$$ and $$\displaystyle{\left|{A}{D}\right|}={12}$$
Since C is unit circle, We have $$\displaystyle{\left|{A}{C}\right|}={1}$$
Plugging the values for $$\displaystyle{\left|{A}{D}\right|}$$ and $$\displaystyle{\left|{A}{C}\right|}$$ into the formula gives,
$$\displaystyle{\left|{C}{D}\right|}={3}\sqrt{{{2}}}$$. Since CD is perpendicular to AB .
Therefore, $$\displaystyle{C}{\left({12},\ {3}\sqrt{{{2}}}\right)}$$
$$\displaystyle{{\sin{{60}}}^{{\circ}}=}{3}\sqrt{{{2}}}$$ and $$\displaystyle{{\cos{{60}}}^{{\circ}}=}{12}$$