Question

In \[\Delta ABC,DE||BC\], AD = 5.7 cm, BD = 9.5 cm, EC = 6 cm. Find AE.

Answer 1

Hint: To solve this question, we will first prove that the triangles ADE and ABC are similar triangles. After proving these triangles similar, we will make use of the fact that if two triangles are similar, then the ratio of their corresponding sides are equal. Then with the help of this, we will find the length of AE.

Complete step-by-step answer:
Before, finding the length of AE, we will prove that the triangle ADE is similar to the triangle ABC. For this, we will determine the relationship between \[\angle ABC\text{ and }\angle ADE\]. Let us take \[\angle ABC=\alpha ;\angle ACB=\beta .\]

Now, here we can see that DE is parallel to BC and AB is transversal, so we have the sum of the interior angles as \[{{180}^{o}}.\] Thus, we get the following relation:
\[\angle ABC+\angle BDE={{180}^{o}}\]
\[\alpha +\angle BDE={{180}^{o}}\]
\[\angle BDE={{180}^{o}}-\alpha .....\left( i \right)\]
Now, the sum of the angles \[\angle ADE\text{ and }\angle BDE\] will be \[{{180}^{o}}\] because AB is a straight line and the angle of the straight line is \[{{180}^{o}}.\] Thus, we get the following relation:
\[\angle ADE+\angle BDE={{180}^{o}}.....\left( ii \right)\]
From equations (i) and (ii), we have,
\[\angle ADE+{{180}^{o}}-\alpha ={{180}^{o}}\]
\[\angle ADE=\alpha \]
Similarly, we will assume that \[\angle ACB=\beta \]. Now, by similar calculations done above, we can say that \[\angle AED\] will also be equal to \[\beta .\] Thus, we get,
\[\angle AED=\beta \]
Now, we will consider the triangle ADE and ABC. In these triangles, we can see that,
\[\angle ADE=\angle ABC=\alpha \]
\[\angle AED=\angle ACB=\beta \]
\[\angle BAC=\angle DAE\left[ \text{Common Angles} \right]\]
Thus, we have,
\[\angle ADE=\angle ABC\]
\[\angle AED=\angle ACB\]
\[\angle BAC=\angle DAE\]
Here, we can see that the angles of the corresponding triangles are equal. So, here we are going to the AAA similarity theorem. According to this theorem, if all the angles of the corresponding triangles are equal, then the triangles will be similar to each other. Thus, triangles ABC and ADE are similar triangles. Now, we know that if two triangles are similar, then the ratios of their corresponding sides are equal. Thus, in \[\Delta ABC\text{ and }\Delta ADE\], we get,
\[\dfrac{AD}{AB}=\dfrac{AE}{AC}\]
\[\Rightarrow \dfrac{AD}{AD+AB}=\dfrac{AE}{AE+AC}\]
\[\Rightarrow \dfrac{5.7}{5.7+9.5}=\dfrac{AE}{AE+6}\]
\[\Rightarrow \dfrac{5.7}{15.2}=\dfrac{AE}{AE+6}\]
\[\Rightarrow 5.7AE+6\left( 5.7 \right)=15.2AE\]
\[\Rightarrow 15.2AE-5.7AE=6\left( 5.7 \right)\]
\[\Rightarrow 9.5AE=34.2\]
\[\Rightarrow AE=\dfrac{34.2}{9.5}\]
AE = 3.6 cm

Note: We can also solve the above question as shown: First, we will construct a line EF parallel to DB. Now, EFDB will be a parallelogram, so EF = DB = 9.5 cm. Instead of proving the triangle ADE and ABC as similar and obtaining the ratios, we can also prove \[\Delta ADE\text{ and }\Delta EFC\] as similar. After these, we will prove the ratios as equal, thus we have:
\[\dfrac{AD}{EF}=\dfrac{AE}{EC}\]
\[\dfrac{5.7}{9.5}=\dfrac{AE}{6}\]
\[AE=\dfrac{6\times 5.7}{9.5}\]
AE = 3.6 cm