Length Ratio

$ABCD$ be a parallelogram and $AC$ as diagonal firstly through $A$ , we will draw a line parallel to $DE$ . Let it be $AZ$ and it will meet the extended $CD$ at $Z$ . Now $AEDZ$ is a parallelogram. $AE$ is equal to $ZD$ . let $AE$ be 'x'. $ZD$ is also x, $EB$ is 234x. so, $AB$ is $235x$ . $AB$ is equal to $CD$ so $CD$ is $235x$ . also $ZC$ is $ZD+DC$ which is $x +235x =236x$ . In triangle $ACZ$ by thales theoram, $\frac {AC}{AF}$ is equal to $\frac {ZC}{ZD}$ so $frac {AC}{AF}$ is equal to $\frac {236x}{x} = 236$ .

Draw BD and let it intersect AC at Q. We thus know that AQ = QC and BQ =QD. Also, by menelaus, we have that AE/EB * BD/DQ * QF/FA = 1. Thus 1/234 2 QF/FA =1, or QF/FA = 118. Thus AC/AF = 2 (QF+FA)/FA = 2+2 117 = 236.

Let AE = x and EB = 234x. Then, DC = AB = AE + EB = 235x. Notice that triangle AEF and triangle CDF are similar, so,

AE / AF = DC / CF x / AF = 235 x / CF CF / AF = 235.

But we need to find AC / AF, or equivelantly as

(AF + CF)/AF = 1 + CF/AF = 1 + 235 = 236

Let the length of BE be 234x, then AE = x. Hence AB = DC = 235x. Triangle AEF is similar to triangle DCF, since angle AEF = angle DCF, angle AFE = angle DFC, and angle FAE = angle FDC. Since AE = x and DC = 235x, then AE : DC = AF : FC = 1 : 235, thus AC = 235 + 1 = 236 units. Thus AC : AF = 236

suppose AE = x, EB = 234x. (AB=235x) AF : FC = 1 : 235 ; FC : AF = 235 AC = AF + FC AC : AF = (AF + FC) : AF = 1 + 235 = 236 So, AC : AF is 236

we have EB=234 AE, so AB=235 AE,i.e., DC=235 AE. triangles AFE and FCD are similar, because $\angle AFE$ = $\angle FCD$ and $\angle FAE$ equals $\angle FCD$ , being alternate. So AF/FC = AE/DC=1/235, so FC=235 AF, so AC=236*AF. so the required ratio is 236.

Let $AE = x$ . Then $EB = 234x$ .

Since $AB \parallel DC$ , $\angle EAC = \angle ACD$ and $\angle AED = \angle EDC \Rightarrow$ triangles $FAE$ and $FCD$ are similar. Thus, $\frac{DC}{AE} = \frac{FC}{AF}$ . But $DC = AB = 235x$ (because $ABCD$ is a parallelogram), so $\frac{DC}{AE} = \frac{235}{1} = \frac{FC}{AF} \Rightarrow 235\times AF = FC \Rightarrow$ $AC : AF = (AF + FC) : AF = 236AF : AF = \boxed{236}$

Construct Line EG parallel to AD such that G is on AC. Then AG:CG=1:234 and then EG:AD=1:235. So AF:FG=235:1. Then AF:FG:GC=235:1:234*235. Then AC:AF=236.

AF=x, FC=234x+x, AC=236x

just clearly draw the perfect picture at first... you have drawn the following straight lines :DE,AC. now we know from the parallelogram ABCD that AB||DC and the straight line DE is sector so angle(AED)=angle(EDC) [alternate angle]and likewise when AC is the sector then angle(BAC)=angle(ACD).[alternate angle]....... now taking the two triangle AEF and triangle DFC::we get,,, 1)angle(AEF)=angle(FDC).2)angle(EAF)=angle(FCD). 3)angle (AFE)=angle(DFC)[vertically opposite].....so triangle AEF and triangle DFC are equal...hence [AE/DC]=[AF/FC] =>[DC/AE]=[FC/AF] =>[DC/AE]+1=[(FC+AF)/AF]=[AC/AF] =>[AC/AF]=[234 (DC/EB)]+1 Now AE/EB=1/234 So (AE+EB)/EB=1+(1/234) =>AB/EB=235/234 =>DC/EB=235/234 SO COME TO PREVIOUS RESULT:::[AC/AF]=[234 (DC/EB)]+1 =[234*(235/234)]+1 =235+1=236
here comes the answer,,,, thanking you ,,,,,,,,,,,,,/-

Since $AB \parallel CD$ , we have: $\angle EAF = \angle FCD$ and $\angle AEF = \angle FDC$ . By opposite angles, $\angle EFA = \angle CFD$ . Thus, by angle-angle-angle, triangles $FAE$ and $FCD$ are similar. Therefore, $\frac{FC}{FA} = \frac{CD}{AE} = \frac{AB}{AE}$ , where the final equality follows since $ABCD$ is a parallelogram and $AB = CD$ .

So we have, $AC = AF + FC \Rightarrow \frac{AC}{AF} = 1 + \frac{FC}{AF}\$ . Substituting the above relation for $\frac{FC}{AF}$ , we have: $\frac{AC}{AF} = 1 + \frac{AB}{AE}$ $= 1 + \frac{AE + EB}{AE}$ $= 1 + 1 + \frac{EB}{AE}$ $= 2 + 234 = 236$ .