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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.

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Yes there is, drop a perpendicular down from $E$ to $AB$ and denote the point as $X$. Let $EX = x$ and $AX = 12-y$ and $BX = y$. Since $\triangle EBX $ ~ $\triangle DBA$, we have that $\frac9x = \frac{12}y$. Also, since we have $\triangle CBA$ ~ $\triangle EXA$, we have that $\frac5x = \frac{12}{12-y}$. Now, solving for $x,y$ using our equations, we get $x=\frac{45}{14}, y = \frac{30}{7}$. Plugging in these values, we get $AX = \frac{54}{7}$ and $BX = \frac{30}{7}$ and $EX = \frac{45}{14}$. Now using the pythagorean theorem, we get $AE = \frac{117}{14}$ and $BE=\frac{75}{14}$. Now, adding these lengths, we get that the sum is $\frac{96}{7}$

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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, $\boxed{AE + BE= x + y = \frac{96}{7}.}$

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  • $\begingroup$ very nice solution $\endgroup$
    – Some Guy
    Feb 16 at 3:18

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