Question

In a triangle $\mathrm{\Delta }ABC$, $DE\parallel BC$ and $CD\parallel EF$ . Prove that $A{D}^2=AF\times AB$ .

Answer 1

Hint: For solving this question, first we will see the result of the Basic Proportionality theorem. After that, we will use it for $\mathrm{\Delta }ABC$, $\mathrm{\Delta }ADC$. Then, we will solve accordingly to prove the desired result easily.

Complete step-by-step solution -
Given:
It is given that, there is $\mathrm{\Delta }ABC$, $DE\parallel BC$ and $CD\parallel EF$ for the following figure:

To Prove: We have to prove that, $A{D}^2=AF\times AB$.
Now, before proceeding we should understand an important theorem which is called the Basic Proportionality theorem.
Basic Proportionality Theorem $\left(BPT\right)$ :
Consider a $\mathrm{\Delta }ABC$ and a line PQ parallel to the side BC intersects side AB at Q and side AC at P. For more clarity look at the figure given below:

Now, as PQ is parallel to the side BC so, $\mathrm{\angle }AQP=\mathrm{\angle }ABC$ and $\mathrm{\angle }APQ=\mathrm{\angle }ACB$ as they are corresponding angles.
Now, consider $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }AQP$ . Then,
$\begin{array}{rl}& \mathrm{\angle }BAC=\mathrm{\angle }QAP\left(common\right)\\ & \mathrm{\angle }AQP=\mathrm{\angle }ABC\left(correspondingangles\right)\\ & \mathrm{\angle }APQ=\mathrm{\angle }ACB\left(correspondingangles\right)\end{array}$
Now, as all the angles of the triangles are equal so, $\mathrm{\Delta }ABC\sim \mathrm{\Delta }AQP$ . And we know that for every similar triangles ratio of the corresponding sides is equal. Then,
${\displaystyle \frac{AB}{AQ}}={\displaystyle \frac{AC}{AP}}={\displaystyle \frac{QP}{BC}}$
Now, from the Basic Proportionality theorem, we say that if PQ is drawn parallel to the side BC and it intersects side AB at Q and side AC at P. Then,
${\displaystyle \frac{AB}{AQ}}={\displaystyle \frac{AC}{AP}}={\displaystyle \frac{QP}{BC}}$
Now, we come back to our problem in which we have the following figure:

Now, we consider $\mathrm{\Delta }ABC$ in which $DE\parallel BC$. Then,
${\displaystyle \frac{AD}{AB}}={\displaystyle \frac{AE}{AC}}(BPT)..................\left(1\right)$
Now, we consider $\mathrm{\Delta }ADC$ in which $CD\parallel EF$. Then,
${\displaystyle \frac{AF}{AD}}={\displaystyle \frac{AE}{AC}}(BPT)..................\left(2\right)$
Now, we will equate the equation (1) and (2). Then,
$\begin{array}{rl}& {\displaystyle \frac{AD}{AB}}={\displaystyle \frac{AE}{AC}}={\displaystyle \frac{AF}{AD}}\\ & \Rightarrow {\displaystyle \frac{AD}{AB}}={\displaystyle \frac{AF}{AD}}\\ & \Rightarrow AD\times AD=AF\times AB\\ & \Rightarrow A{D}^2=AF\times AB\end{array}$
Now, from the above result, we conclude that for the given triangle $A{D}^2=AF\times AB$.
Hence, proved.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct answer quickly. After that, we should not be confused while applying the result of the Basic proportionality theorem, and we should proceed stepwise to avoid confusion. Moreover, while solving such questions, we should always refer to the diagram if we get stuck at some step and try to figure out the best way to solve it.