In triangle $ABC$, points $D$ and $F$ are on $\overline{AB},$ and $E$ is on $\overline{AC}$ such that $\overline{DE}\parallel \overline{BC}$ and $\overline{EF}\parallel \overline{CD}$. If $CE =3$ and $DF = 3$, then what is $BD$?
If CE = 3 and DF = 3, then what is BD?
Im Dreieck ABC liegen die Punkte D und F auf {AB} und E auf {AC}, sodass {DE}\parallel {BC} und {EF} parallel {CD} sind. Wenn CE = 3 und DF = 3, was ist dann BD?
Let AF = AE = c and BD = x.
Then, according to the first ray theorem:
s+3s=s+3+xs+3s2+6x+9=s2+3s+sxsx=3s+9x=3+9s
To solve the question, it is necessary to specify an additional value, for example AE.
!