enter image description here

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.


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 $\frac{96}{7}$

enter image description here


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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.