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enter image description here Ok so I have this problem in which we have to find out the length AB by using the limited data provided. How can we solve this? Do we use the congruent angles property of two parallel lines being intersected by a traversal or do we something completely different? The answer is 8 I would really appreciate if you solve it step b step :)

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Since $AF//ED$, we have that $\frac{EC}{CF}=\frac{DC}{CA} \Rightarrow \frac{18}{24}=\frac{12}{CA} \Rightarrow CA=16\ (1).$

Now since $GB//EF$, we have that $GB//FC$. Thus in the triangle $ACF$ we get that $\frac{AG}{GF}=\frac{AB}{BC}$ and so $AB=BC\ (2)$, because it is given that $AG=GF$.

Finally we combine $(1)$ and $(2)$, we have $16=CA=AB+BC=2AB\Rightarrow AB=8$.

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Hints: Triangles $ABG$, $ACF$, and $DCE$ are all similar to each other.

  • Since $AG = GF$, we know the scale factor between $ABG$ and $ACF$, so we can use $CF = 24$ to determine the length of $BG$.
  • Combining this with $EC = 18$, we can determine the scale factor between $ABG$ and $DCE$, so we can use $CD = 12$ to determine the length of $AB$.
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No properties of congruent triangles are needed to solve this problem. Just good old-fashioned parallel line and similarity geometry.

You need to know that

  • corresponding or alternate angles on parallel lines are equal,

  • similar triangles have equal angles and

  • the corresponding sides of similar triangles are in proportion.

In order to find AB given the lengths of sides of some other triangles, we need to use ratios of similar triangles involving the length AB. So we need first to prove that $\triangle{ABG} ||| \triangle{AFC}$.

To prove this,

$GB||EF$ (given) so $\angle{GBA}=\angle{FCA}$ and $\angle{AGB}=\angle{AFC}$ (corresponding angles on parallel lines.)

In $\triangle{ABG}$ and $\triangle{AFC}$

  • $\angle{GBA}=\angle{FCA}$ (proved)
  • $\angle{AGB}=\angle{AFC}$ (proved)
  • $\angle{GAB}=\angle{FAC}$ (common)

$\therefore \triangle{ABG} ||| \triangle{AFC}$ (equiangular.)

Now we are also given that $AG = GF$ i.e. $AF = 2AG$.

We proved that the triangles are similar so

$\displaystyle\frac{AG}{AF}=\frac{GB}{CF}$ which means $\displaystyle\frac{AG}{2AG}=\frac{GB}{24}$ and hence $GB=12$.

Similarly (excuse the pun) we now know that $\displaystyle\frac{AB}{AC}=\frac{GB}{FC}$ so we must find the length $AC$ to find $AB$.

Now, we are given $AF||ED$, so $\angle{FAC}=\angle{EDC}$ and $\angle{CFA}=\angle{CED}$ (alternate angles on parallel lines.) $\angle{FCA}=\angle{DCE}$ since they are vertically opposite.

Using these facts we say that $\triangle{DCE} ||| \triangle{ACF}$ and so $\displaystyle\frac{AC}{CF}=\frac{DC}{CE}$

$\therefore\displaystyle\frac{AC}{24}=\frac{12}{18}$ and $AC=16$.

Since $\displaystyle\frac{AB}{AC}=\frac{GB}{FC}$, $\displaystyle\frac{AB}{16}=\frac{12}{24}$ and $AB=8$.