6. Equilateral triangles $A B C$ and $C D E$ are congruent triangles. If $|\measuredangle A C E|=74^{\circ} 30^{\prime}$, what is the measure of $\measuredangle A B E$ ?

![](https://cdn.mathpix.com/cropped/2024_06_03_a5588eb755b4ad428ffbg-2.jpg?height=454&width=651&top_left_y=630&top_left_x=605)
First method:

![](https://cdn.mathpix.com/cropped/2024_05_30_b1442ac0d12d1cd02522g-2.jpg?height=429&width=608&top_left_y=2264&top_left_x=313)

Triangle $BCE$ is an isosceles triangle because $|BC| = |CE|$, which is a consequence of the congruence of the equilateral triangles $ABC$ and $CDE$.

The measure of each interior angle of the equilateral triangles is $60^{\circ}$, so it follows that $|\measuredangle BCE| = |\measuredangle BCA| + |\measuredangle ACE| = 60^{\circ} + 74^{\circ} 30' = 134^{\circ} 30'$.

$\overline{BE}$ is the base of the isosceles triangle $BCE$, so it follows that

$|\measuredangle EBC| = |\measuredangle CEB| = \left(180^{\circ} - 134^{\circ} 30'\right) : 2 = 45^{\circ} 30' : 2 = 44^{\circ} 90' : 2 = 22^{\circ} 45'$.

Further, it follows that $|\measuredangle ABE| = |\measuredangle ABC| - |\measuredangle EBC| = 60^{\circ} - 22^{\circ} 45' = 37^{\circ} 15'$.