Question

In the given figure, GC $\parallel $ BD and GE $\parallel $ BF If AC = 3 cm and CD = 7 cm, then find the value of $\dfrac{AE}{AF}$

Answer 1

Hint: As we can see in the figure, there are 4 triangles viz, triangle ACG, triangle ADB, triangle AEG and triangle AFB in the bigger kite AFBD. We know that the corresponding sides of similar triangles are in ratio. First, we will prove that triangle ACG is similar to triangle ADB. We can thus prove ratio of AG to AB is same as that of AC and AD. Then we will prove that triangle AEG is similar to triangle AFB. Then, we can show that ratio of AE to AF is same as that of AG to AB.

Complete step by step answer:
It is given that AC = 3 and CD = 7. From the figure, we can see that AD = AC + CD
$\Rightarrow $ AD = 3 + 7 = 10 cm.
Therefore, length of AD is 10 cm.
Now, we will consider triangle AGC and triangle ABD.
$\angle $A = $\angle $A … [common angles]
It is given that GC $\parallel $ BD. This means AD acts as a transverse of parallel lines CG and BD.
$\angle $ACG = $\angle $ADB … [ corresponding angles of parallel lines]
Therefore, from AA rule, we can say that triangle ACG is similar to triangle ADB.
$\begin{align}
  & \Rightarrow \dfrac{AC}{AD}=\dfrac{CG}{DB}=\dfrac{AG}{AB} \\
 & \Rightarrow \dfrac{AG}{AB}=\dfrac{3}{10} \\
\end{align}$
Now, we shall consider triangle AGE and triangle ABF.
$\angle $A = $\angle $A … [common angles]
It is given that GE $\parallel $ BF. This means AF acts as a transverse of parallel lines CE and BF.
$\angle $AEG= $\angle $AFB … [corresponding angles of parallel lines]
Therefore, from AA rule, we can say that triangle ACG is similar to triangle ADB.
$\begin{align}
  & \Rightarrow \dfrac{AE}{AF}=\dfrac{EG}{FB}=\dfrac{AG}{AB} \\
 & \Rightarrow \dfrac{AE}{AF}=\dfrac{3}{10} \\
\end{align}$

Therefore, ratio of AE to AF is 3 to 10.

Note: Students can remember the result that if a line is drawn intersecting two sides of a triangle and is parallel to the third side of the triangle, then the two triangles formed are similar to each other.