Find the length of $AE+EB$ ?
(A) $\frac{128}{7}$
(B) $\frac{112}{7}$
(C) $\frac{100}{7}$
(D) $\frac{96}{7}$
(E) $\frac{56}{7}$
My solution:
For $\Delta AEB$ :
$\angle BAC = sin^{-1}(\frac{5}{13}) = 22.62^{\circ}$
$\angle ABD = sin^{-1}(\frac{9}{15}) = 36.87^{\circ}$
$\angle AEB = 180 – \angle BAC – \angle ABD = 120.52^{\circ}$
Use Law of Sines:
$\frac{AE}{sin(\angle ABD)} = \frac{AB}{sin(\angle AEB)}$
$AE = \frac{AB}{sin(\angle AEB)} \times sin(\angle ABD)$
$AE = \frac{12 \times \frac{9}{15}}{sin 120.51} = 8,372$
$\frac{EB}{sin(\angle BAC)} = \frac{AB}{sin(\angle AEB)}$
$EB = \frac{AB}{sin(\angle AEB)} \times sin(\angle BAC)$
$EB = \frac{12 \times \frac{5}{13}}{sin 120.51} = 5.367$
$AE+EB = 8.372 + 5.367 = 13,7387 \approx \frac{96}{7}$ Answer: (D)
My question:
Is there a solution without having to compute both arcsin $\angle BAC$ & $\angle ABD$? The reason I'm asking is because the choices all have 7 as denominators. So I'm guessing there may be a solution that contains only integers.
Best Answer
Note that $\triangle AED\sim\triangle CEB$. Then, letting $AE = x$ and $BE = y$, we have:
$$DE = \frac{9}{5}y,\ CE = \frac{5}{9}x$$
From the similar triangles. Now, we must have $AE + CE = AC$ and $BE + DE = BD$. Noting that $AC = 13$ and $BD = 15$:
$$x + \frac{5}{9}x = 13$$
$$y + \frac{9}{5}y = 15$$
Solving, we find $\displaystyle x = \frac{117}{14}$ and $\displaystyle y = \frac{75}{14}$. Thus, $\boxed{AE + BE= x + y = \frac{96}{7}.}$