Find the Distance

Let the foot of the perpendicular from $D$ to $\overline{AB}$ be $G$ . Since $\triangle ABC\sim \triangle AGD$ , we get that $\overline{AG}=\frac{15}{17}$ and $\overline{DG}=\frac{8}{17}$ . This means that $\overline{GE}=\overline{AE}-\overline{AG}=3-\frac{15}{17}=\frac{36}{17}$ . Also, $\overline{BE}=\overline{AB}-\overline{AE}=15-3=12$ . Now, noting that $\triangle GDE\sim \triangle BFE$ , we get that $\frac{\overline{DG}}{\overline{GE}}=\frac{\overline{BF}}{\overline{BE}}$ , and using $\overline{DG}=\frac{8}{17}$ , $\overline{GE}=\frac{36}{17}$ , and $\overline{BE}=12$ , we get that $\overline{BF}=\frac 83$ , giving an answer of $8+3=\boxed{11}$ .

Drop the altitude from $D$ to $\overline{AE}$ and have it intersect at $G$ . Now, note that $\Delta ABC \sim \Delta AGD$ since $\angle A$ is shared and $\angle ABC = \angle AGD = 90^\circ$ . Also note that $\Delta GED \sim \Delta BEF$ since $\angle BEF$ and $\angle GED$ are vertically opposite and $\angle DGE = \angle FBE = 90^\circ$ .

Since $\Delta ABC \sim \Delta AGD$ , $\frac{AG}{AD} = \frac{AB}{AC} = \frac{15}{17}$ so $\overline{AG} = 1\times \frac{15}{17}$ and $\overline{EG} = \overline{AE} - \overline{AG} = 3 - \frac{15}{17} = \frac{36}{17}$ .

Since $\Delta GED \sim \Delta BEF$ , $\frac{BF}{BE} = \frac{DG}{EG} = \frac{8/17}{36/17}$ and $\overline{BE} = \overline{AB} - \overline{AE} = 15 - 3 = 12$ so $\overline{BF} = 12 \times \frac{8}{36} = \frac{8}{3}$ .

Our answer is $8 + 3 = \boxed{11}$ .

Begin by drawing $\Delta ABC$ on a two-dimensional Cartesian plane with $A(0,15)$ , $B(0,0)$ and $C(8,0)$ .

Next, determine the coordinates of $D$ and $E$ .

The ratio of $\overline{AD}$ to $\overline{DC}$ is $\frac{1}{17}$ . Let $G$ be the foot of the perpendicular from $D$ to the x-axis. Since $\Delta DGC$ is similar to $\Delta ABC$ , the x-coordinate of $D$ is $\frac{1}{17}\overline{BC} = \frac{8}{17}$ and the y-coordinate of $D$ is $(1-\frac{1}{17})\overline{AB}=\frac{240}{17}$

$\Rightarrow D(\frac{8}{17},\frac{240}{17})$ .

$E$ is $3$ units below $A$ and its coordinates are obtained trivially.

$\Rightarrow E(0,12)$

With the coordinates of $D$ and $E$ , determine the equation of the line $\overleftrightarrow{DE}$ .

$y-\frac{240}{17} = \frac{\frac{240}{17}-12}{\frac{8}{17}-0}(x-\frac{8}{17})$

$\Rightarrow \overleftrightarrow{DE}:y=\frac{9}{2}x+12$

The line $\overleftrightarrow{BC}$ lies along the x-axis, so the point of intersection of $\overleftrightarrow{DE}$ and $\overleftrightarrow{BC}$ , $F$ , is also the x-intercept of $\overleftrightarrow{DE}$ . Substitute $y=0$ into the equation of $\overleftrightarrow{DE}$ and solve for $x$ to find $F$ .

$0=\frac{9}{2}x+12$

$\Rightarrow F(-\frac{8}{3},0)$

Drop the negative to obtain the distance $BF$ as $\frac{8}{3}$ , which is in the form $\frac{a}{b}$ .

Therefore $a+b=8+3=11$ .

The solution involves three triangles to solve the problem: the biggest one $\bigtriangleup ABC$ , and the two triangles formed by the two lines $\overleftrightarrow{DE}$ and $\overleftrightarrow{BC}$ : $\bigtriangleup DAE$ and $\bigtriangleup EBF$ .

$1.$ Finding $\angle CAB$ via rt. $\bigtriangleup ABC$ using simple trigonometry, which is:

$m\angle CAB = \cos^{-1} (\frac {15}{17})$

$2.$ Use Cosine Law to find $\overline {DE}$ in $\bigtriangleup DAE$ :

We know that $m\angle CAB = \cos^{-1} (\frac {15}{17}) = m\angle DAE$ . We first write the cosine law equation:

$DE^2 = DA^2 + AE^2 -2(AD)(AE) \cos \angle DAE$ . Substituting the given, DA = 1, and AE = 3:

$DE^2 = 1^2 + 3^2 -2(1)(3) \cos (\cos^{-1} (\frac {15}{17}))$ .

We find that:

$DE= \sqrt {\frac {80}{17}} = \frac {4\sqrt{85}}{17}$ .

$3.$ Find $m\angle AED$ using Sine Law:

$\frac {\overline{DE}}{\sin \angle DAE} = \frac {\overline{DA}}{\sin \angle DEA}$

By substituting the values we arrive at:

$\frac {(\frac {4\sqrt{85}}{17})}{\sin (\cos^{-1} (\frac {15}{17}))} = \frac {1}{\sin \angle AED }$

By performing MPE:

$\sin \angle AED = \frac {2}{\sqrt{85}}$

Therefore:

$m\angle AED = \sin^{-1} \frac {2}{\sqrt{85}}$ .

$4.$ Use simple trigonometry to find $\overline {BF}$ in right $\bigtriangleup EBF$ . Note: $\angle BEF = \angle AED$ by VAC theorem; and $\overline {EB} = 15 - 3 = 12$

$\overline {BF} = \overline {EB} \tan ( \angle BEF)$

By substitution:

$\overline {BF} = (12) \tan ( \sin^{-1} \frac {2}{\sqrt{85}})$

Using the calculator:

$\overline {BF} = \frac {8}{3} = \frac {a}{b}$

Therefore:

$a + b = 8 + 3 = 11$ .

The points $D$ , $E$ and $F$ are colinear and the Theorem of Menelaus tells us that $\frac{AE}{EB} \cdot \frac{BF}{FC} \cdot \frac{CD}{DA} = 1.$ Noting that $AE = 3$ , $EB = 12$ , $CD = 16$ , $DA = 1$ , and $FC = BF + 8$ gives us the equation $\frac{3}{12} \cdot \frac{BF}{BF + 8} \cdot \frac{16}{1} = 1,$ from which we get $\frac{BF}{BF+8} = \frac{1}{4}.$ Clearing fractions leads to $4 BF = BF + 8$ and thus $BF = \frac{8}{3}$ . So $\frac{a}{b} = \frac{8}{3}$ and $a + b =$ 11.

Let A, B, and C be at points (0, 15), (0, 0), and (8, 0), respectively. Then E is at (0, 12). Drawing a perpendicular from D to AB at G, we form similar triangles, so we have AG/AB = AD/AC = GD/BC. Solving, we get AG = 15/17 and DG = 8/17, so point D is at (8/17, 240/17).

The line through D and E goes through the points (0, 12) and (8/17, 240/17), so it has the form y = 9x/2 + 12. Since segment BC is on line y = 0, F is at the x-intercept of line DE; plugging in y = 0 gives x = -24/9, or -8/3. Then BF is also -8/3 since B is at the origin. Our answer is 8 + 3 = 11.

1. find angle CAB using trigonometric functions. Angle CAB = 28.07 degrees.
2. find segment DE by cosine law. (DE)^2 = 3^2 + 1^2 - 2 3 1*cos(28.07) DE = 2.169 units
3. find angle DEA by sine law. sin (DEA) = (sin (28.07))/2.169 angle DEA = 12.53 degrees
4. angle BEF = angle DEA (vertical angles are congruent) angle BEF = 12.53 degrees
5. BE = BA - EA = 15 - 3 = 12 units
6. to find segment BF in triangle BFE, use trigonometric functions. segment BF = BE tan (BEF) = 12 tan (12.53)
7. segment BF = a/b = 8/3 units
8. a+b = 8+3 = 11 units

All of you had made a diagram so I will continue from there.We are given a point D on AC, draw a perpendicular from D to any Point on BC,such that it become parallel to side AB.Let that point on BC be denoted by X.We know that BC=8, so let BX=p, so XC=8-p. Let BF=y.Now Applying the concept of similarity in ABC, we get AD/BX = CD/XC. Putting the values we get p=8/17., so 8-p=128/17.Then applying Pythagoras theorem in DXC, we get the perpendicular which we dropped from D equals to 240/17=DX .Now again applying the concept of similarity in FDX, we get FB/FX=EB/DX., Hence putting the values we get y=3/8.

Call point B the origin $(0,0)$ so A is $(15,0)$ and C is $(0,8)$. Then E would be $(12,0)$. Now drop a perpendicular from D to AB. We see that it creates a new triangle similar to ABC with hypotenuse 1. Thus, D would be $(15-\frac{15}{17},{8}{17})$. Now we find the equation of the line that contains D and E. the absolute value of the y- intercept would be our answer. The slope is $\frac{2}{9}$. Plug in $(12,0)$ to get b=$\frac{-8}{3}$ so the distance is $\frac{8}{3}$ and $8+3=\boxed{11}$.