Distance Around a Parallelogram

Since $AD \parallel BC$ , $\vartriangle ADG \sim \vartriangle BFG$ by AA similarity.

Likewise, $\vartriangle ABF \sim \vartriangle EDA$

$AG:GF=180:108=5:3$ , thus by similarity, $AD:BF=5:3$ and $AE:AF=5:3$

$AF=288$ , so $AE=480$

$EF=AE-AF=480-288=192$

tr.BGF ~ tr.DGA

So, GF/GA=BF/AD 108/180=BF/AD BF=3/5AD

So, BF=3/5BC Also, CF=2/5BC

CF/BF=2/3

tr.FEC ~ tr.FAB

FE/FA=CF/BF FE/288=2/3 FE=192 So, EF =192

Due to the fact $AB \parallel DE$ , we have:

$\angle BAG = \angle DEG$ , $\angle ABG = \angle GDE$ , which implies $\triangle ABG \sim \triangle EDG$ .

$\angle ABF = \angle ECF$ , $\angle BAF = \angle CEF$ , which implies $\triangle ABF \sim \triangle ECF$ .

Therefore $\frac{EG}{AG}=\frac{ED}{AB}$ and $\frac{EF}{AF}=\frac{EC}{AB}$

Note $EG=108+EF$ , $AG=180$ , $AF=180+108=288$ and $ED=EC+CD=EC+AB$ .

Thus, $\frac{108+EF}{180}=\frac{EC+AB}{AB}$ and $\frac{EF}{288}=\frac{EC}{AB}$ .

Subtracting 1 from the first equation, we get: $\frac{108-180+EF}{180}=\frac{EC}{AB}$

Replacing $\frac{EC}{AB}$ with $\frac{EF}{288}$ for that equation, we get: $\frac{EF-72}{180}=\frac{EF}{288}$

$(EF-72) \times 288=EF \times 180$

$EF \times 288-20736=EF \times 180$

$EF \times (288-180)=20736$

$EF \times 108=20736$

Thus $EF=\frac{20736}{108}=192$ .

Let $x$ be the length of side $\overline{EF}$. Since triangles $\Delta ECF$ and $\Delta EDA$ are similar, we see that $\frac{x}{108+180+x}=C$ where $C$ is the constant of proportionality between the two triangles. Let $y$ be the length $\overline{AD}$, and let $t$ be the length $\overline{BF}$, and we see that we also have that $\frac{y-t}{y}=C$. Using the similarity of the triangles $\Delta GBF$ and $\Delta GDA$, we see that $\frac{t}{y}=\frac{108}{180}$, and so $C=1-\frac{108}{180}=\frac{72}{180}$. Solving the equation $\frac{x}{288+x}=\frac{72}{180}$, we obtain the final answer $x=192$.

Let EF be $x$ . Let AB be $a$ and CE be $b$ .

Since triangle CEF and triangle BAF are similar, $\frac {x} {180+108} = \frac {b} {a}$ .

Since triangle ABG and triangle EDG are similar, $\frac {x+108} {180} = \frac {a+b} {a} = 1 + \frac {b} {a}$ .

Therefore, from the 2 equations, we get:

$\frac {x} {288} + 1 = \frac {x+108} {180}$

So, $180x + 180 \cdot 288 = 288x + 108 \cdot 288$

$108x = 72 \cdot 288$

$x = 192$ and we are done.

from the given picture we see that triangle DEG & triangle AGB are similar. therefore GE/DE = AG/AB ..........(1). from triangle ADE & triangle FCE we get AE/FE = DE/CE ............(2).(since triangle ADE & triangle FCE are similar.) from triangle ABF & CFE we get AF/FE = AB/CE ...........(3).(since triangle ABF & triangle CFE are similar.) now from (2) we get DE = AE x CE/FE . from (1) we get FExGE/AExCE = AG/AB (putting the value of DE in (1).) therefore AB/CE = AExAG/FExGE . from (3) we get AF/FE = AExAG/FExGE . therefore AF/AG = AE/GE. therefore AFxGE = AExAG or, AFx(GF+FE) = (AF+FE)xAG or, AFxGF+AFxFE = AFxAG+FExAG . or, AFxFE - FExAG = AFxAG - AFxGF or, FEx(AF-AG) = AFx(AG-GF) or, FExGF = (AG+GF)x(AG-GF) or, FE= (AG^2 - GF^2)/GF since AG = 180 , GF=108 therefore, FE =(180^2-108^2)/108=192

$AD \parallel BC$ with transversal $AE \Rightarrow \angle DAE \cong \angle BFA$ (1)

$AB \parallel DE$ with transversal $AE \Rightarrow \angle AED \cong \angle FAB$ (2)

$AD \parallel DE$ with transversal $DB \Rightarrow \angle ADG \cong \angle FBG$ (3) and $\angle EDG \cong \angle ABG$ (4)

1 and 2 $\Rightarrow \triangle DAE \sim \triangle BFA$

1 and 3 $\Rightarrow \triangle DAG \sim \triangle BFG$

2 and 4 $\Rightarrow \triangle DGE \sim \triangle BGA$

By properties of similar triangles: $\frac{AG}{FG} = \frac{EG}{AG}$ $AG=180, FG=108, EG = FG+FE=108+FE$ $\frac{180}{108} = \frac {108+FE}{180} \Rightarrow FE = 192$

A. Triangle ABG is similar to Triangle EDG. Therefore:

AB/AG = ED/EG AB/180 = (DC+EC)/(EF+FG) = (DC+EC)/(EF + 108) AB/180 = (AB + EC)/(EF + 108)

B. Triangle ABF is similar to Triangle ECF. Therefore:

AB/AF = EC/EF AB/288 = EC/EF

C. Triangle ECF is similar to Triangle EDA. Therefore:

EF/EC = EA/ED EF/EC = (EF+ 288)/(EC + CD) = (EF + 288)/ (EC + AB)

To conclude, we have three simultaneous equations relating three lengths:

AB/180 = (AB + EC)/(EF + 108)

AB/288 = EC/EF

EF/EC = (EF + 288)/ (EC + AB)

From here we can solve to obtain the length EF, which is 192 in this case.

As $BD$ and $AF$ are diagonals of the quadrilateral $ABFD$ meeting at $G$ , $\frac{[ABD]}{[FBD]} = \frac{AG}{FG} = \frac{180}{108} = \frac{5}{3}$ Letting $[ABD] = 5x$ , we obtain $[FBD] = 3x$ and $[BCD] = 5x$ so $[FCD] = [BCD] - [FBD] = 2x$ thus $\frac{BF}{FC} = \frac{[FBD]}{[FCD]} = \frac{3}{2}$ As $AB // CE$ , $ABF$ is similar to $CEF$ so $\frac{AF}{EF} = \frac{BF}{FC} = \frac{3}{2}$ and since $AF = AG+GF = 288$ we can calculate that $EF = 192$

AD/FC=AE/EF=(288+EF)/EF and AD/BF=AD/(AD-FC)=AG/GF=180/108

Firstly, you should construct a diagram. Make sure the diagram is accurate, or you will be confused. The next step is to identify some similar triangles to compare ratios. Then, you are done. After drawing the diagram, we can identify that Triangle FEC is similar to Triangle ABF.

So, we claim that FEC is similar to ABF

Proof:

Using the formula AAA( Angle-Angle-Angle), we can prove that FEC is similar to ABF as $\angle CFE=\angle AFB$ because of the opposite angle theorem, $\angle FCE=\angle ABF$ because they are alternate interior angles formed by two parallel lines. Lastly, $\angle FAB=\angle FEC$ because they are also alternate interior angles formed by two parallel lines. QED

Now, we will name line segment AB d, line segment CE z, and line segment FE c.

We will now claim another two triangles are similar,

Claim Triangle AGB is similar to Triangle DGE

Proof:

Using AAA, we can see that $\angle DGE=\angle AGB$ because of the opposite angle theorem, $\angle GED=\angle GAB$ because of the alternate interior angles formed by two parallel lines. Lastly, $\angle GDE=\angle ABG$ because the other angles of the triangles are the same.QED

Now, we can form the two equations that will solve the question.

$\frac{c}{z}=\frac{288}{d}$ (1)

$\frac{d}{d+z}=\frac{180}{108+c}$ (2)

The first equation is quite obvious.

$\frac{c}{z}=\frac{288}{d}$ because of the ratio between Triangle FEC and Triangle AFB.

The second equation is actually not ambiguous too.

$\frac{d}{d+z}=\frac{180}{108+c}$ because of the ratio between Triangle AGB and Triangle DGE.

Usually, an equation with 3 variables is quite impossible to solve without 3 equations. But this counts as a very special case.

By cross-multiplication, we get

$(108+c)d=180(d+z)$ from (2)

We get $cd=288z$ from (1)

By substitution, we get (3)

$3z=2d$ (3)

Then, we substitute (3) into (2),

We will get c=192 and we are done.

By property D in Parallel Lines , triangles $GFB$ and $GAD$ are similar so $\frac {GF}{GA}=\frac {GB} {GD}$ . Likewise, triangles $GED$ and $GAB$ are similar, so $\frac {GE}{GA} = \frac {GD}{GB}$ . Multiplying these equations together, we get $\frac {GF}{GA} \cdot \frac {GE}{GA} = \frac {GB}{GD} \cdot \frac {GD}{GB} =1$ , hence $AG^2 = EG \cdot FG$ . This gives $GE = \frac {AG^2}{FG} = \frac {180 ^2}{ 108} = 300$ . Thus $FE = GE - GF = 300-108 = 192$ .

We can note by AA criteria that $\triangle BGF$ and $\triangle DGA$ are similar, therefore,

$\frac{BF}{DA} = \frac{GF}{GA} = \frac{108}{180} = \frac{3}{5}$

[Note that DA = BC]

Next, $\triangle AFB$ and $\triangle EFC$ are similar, therefore,

$\frac{AF}{EF} = \frac{FB}{FC} = \frac{3}{2}$ [from above ratio of $\frac{3}{5}$ ]

$\frac{180+108}{EF} = \frac{3}{2}$

EF =192

NOTE: It is very important to write the correct corresponding vertices of triangles when mentioning similarity, which makes writing the ratio of sides very easy.

You should not JUST write the name of triangles; but also write them in correct vertices' order!!