Euclidea 4.2

Euclidea 4.2 Angle of $$ 60^{\circ} $$ -2


Construct a straight line through the given point $${\rm A} $$ that makes an angle of $$ 60^{\circ} $$ with the given line.

Goal: 3L, 4E 

Solution 3L



 * 1) Center $$ {\rm A} $$, arbitary radius, draw circle to intersect the line at points $$ {\rm B, C} $$;
 * 2) Center $$ {\rm B} $$, radius $$ \overline{\rm BA} $$, draw circle to intersect $$ \odot {\rm A} $$ at point  $$ {\rm D} $$;
 * 3) Draw angle bisector of $$ \angle DAC $$ to intersect the line at $$ {\rm E} $$, line $$ \overline{\rm AE} $$ is the solution.

Why it works?
This is actually one of the cool and simple solutions. To see how it works, we need one auxiliary line as follows:



From the constructions, a few things are obvious

$$ \angle BAD = 60^{\circ} $$,

$$\angle ABC = \angle ACB $$, and

$$ \angle AEC= \angle ABC + \angle BAE = \angle ACB + \angle BAE $$.

Because of the angle bisector

$$ \angle CAE = \angle DAE = 60^{\circ} + \angle BAE $$

Now consider the triangle $$ \triangle AEC $$, we can write

$$ \begin{align} 180^{\circ}&= \angle ACB + \angle AEC + \angle EAC \\ &=\angle ACB + (\angle ACB + \angle BAE)+ (60^{\circ} + \angle BAE ) \end{align} $$

and this yields

$$ \angle AEC = \angle ACB + \angle BAE = 60^{\circ} $$

Solution 4E



 * 1) Center $$ {\rm A} $$, arbitiary radius, draw circle to intersect the line at points $$ {\rm B, C} $$;
 * 2) Center $$ {\rm B} $$, radius $$ \overline{\rm BA} $$, draw circle to intersect $$ \odot {\rm A} $$ at point  $$ {\rm D} $$;
 * 3) Center $$ {\rm C} $$, radius $$ \overline{\rm CD} $$, draw circle to intersect $$ \odot {\rm B} $$ at point  $$ {\rm F} $$;
 * 4) Connect $$ \overline{\rm AF} $$, to intersect the line at $$ {\rm E} $$, $$ \overline{\rm AE} $$ is the solution.

Why does it work?
The first two steps are exactly the same as the 3L solution. So $$ \odot {\rm C} $$ is the key. This is actually amazing: the triangle $$ \triangle CDF $$ is a equilateral triangle. To see this, we need a few auxiliary lines.



From $$ \odot {\rm A} $$, we have

$$ \angle DCB = \frac{1}{2} \angle DAB = 30^{\circ} $$

Now $$ \angle BCF = 30^{\circ} $$ because $$ \triangle BCF $$ and $$ \triangle BCD $$ are congruent. So $$ \angle DCF = 60^{\circ} $$, and hence the triangle $$ \triangle DCF $$ is an equilateral triangle.

Also notice that the line $$ \overline{\rm CF} $$ passes through point $$ {\rm G}$$, the intersection between $$ \odot A $$ and $$ \odot B $$. This is because $$ \angle BCG = 30^{\circ} $$, also from the property of $$ \odot {\rm A} $$.

With all of these, we have $$ \overline{\rm CF} =\overline{\rm DF} $$. Now triangles $$ \triangle ADF $$ and $$ \triangle ACF $$ are congruent. Recall in Solution 3L we seek the angle bisector of $$ \angle DAC $$, now we have it.

Alternatively, from $$ \odot B $$, we have $$ \angle AFG = \frac{1}{2} \angle ABG = 30^{\circ} $$, this implies $$ \angle AEC =60^{\circ} $$.