Question

In the given figure, $ABC$ and$CEF$ are two triangles, where$BA$ is parallel to $CE$ and $AF:AC=5:8$. Prove that $\Delta ADF~\Delta CEF$. Find AD, if $CE=6cm$. If DE is parallel to BC, then find the ratio of $Area\left(\Delta ADF\right):Area\left(\Delta ABC\right)$

Answer 1

Step 1: Proof for $\Delta ADF~\Delta CFE$:

$\angle DAF=\angle FCE$ (Alternate Angles)

$\angle AFD=\angle CFE$(Vertically opposite angles)

From the above statements, we come to the conclusion that

$\Delta ADF~\Delta CEF$ (By AA axiom)

[AA axiom: In two triangles, if two pairs of corresponding angles are congruent, then the triangles are similar.]

Step 2: Find the value of AD:

We know that, CE = 6 cm.

We proved that $\Delta ADF~\Delta CEF$ and we know the property of similar triangles so

$\frac{AD}{CE}=\frac{AF}{FC}$

From the above figure it is clear that the line AC is formed by two lines AF and FC, so we can write the above equation as :

$\frac{AD}{CE}=\frac{AF}{AC-AF}$

According to the given data we know that $AF:AC=5:8$ and $CE=6cm$

So on substituting the values in above equation we get,

$\frac{AD}{6}=\frac{5}{8-5}$

$\frac{AD}{6}=\frac{5}{3}$

$AD=\frac{6\times 5}{3}=10cm$

Step 3: Find the ratio of $Area\left(\Delta ADF\right):Area\left(\Delta ABC\right)$:

As $DF$ is parallel to $BC$

$\angle D=\angle B$ and $\angle F=\angle C$ (Corresponding angles)

From the above statements, we come to the conclusion that

$\Delta ADF~\Delta ABC$ (By AA axiom)

[AA axiom: In two triangles, if two pairs of corresponding angles are congruent, then the triangles are similar.]

We proved above that $\Delta ABC~\Delta DEC$ and we know the property of similar triangles: "The ratio of the areas of two similar triangles is equal to the square of the ratio of any pair of their corresponding sides".

So putting the theorem into the equation and data we have we get the equation as,

$\frac{Area\left(\Delta ADF\right)}{Area\left(\Delta ABC\right)}=\frac{A{F}^2}{A{C}^2}$

$\frac{Area\left(\Delta ABC\right)}{Area\left(\Delta DEC\right)}=\frac{5^2}{8^2}$

$\frac{Area\left(\Delta ABC\right)}{Area\left(\Delta DEC\right)}=\frac{25}{64}$

So we get the ratio of $Area\left(\Delta ADF\right): Area\left(\Delta ABC\right)= 25:64$.

Thus, we proved that $\Delta ADF~\Delta CEF$, the value of $AD$ is $10cm$ and the ratio of $Area\left(\Delta ADF\right): Area\left(\Delta ABC\right)= 25:64$.