Question

In the given figure, \[\angle CED=\angle CAB\]. Show that \[\Delta CED\sim \Delta CAB\]. Also, find the value of \[x\]?

Answer 1

Hint: In order to solving that \[\Delta CED\sim \Delta CAB\], we first know about the AAA congruence criteria of similarity i.e. if all the three angles of a triangle is corresponding to the angles of the other triangle given in the question. So, if in two triangles, the corresponding angles are equal, i.e., if the two triangles are equiangular, then the two triangles are similar. And to find the value of x, we should know that if two triangles are similar, then their corresponding sides are equal. Then putting the values by using the figure, we would get the value of x.

Complete step by step solution:
Consider the \[\Delta CAB\ and\ \Delta CED\]

\[\Rightarrow \angle CAD=\angle CED\] (It is given)
\[\Rightarrow \angle ACB=\angle ECD\] (Common angle in both the triangle)
So, by AAA criteria of similarity,
\[\Rightarrow \angle CAB\sim \Delta CED\]
\[\therefore \]Hence proved.
It is the property of triangle that if the two triangles are congruent then their corresponding sides are proportional in similar triangles.
So,
\[\Rightarrow \dfrac{CA}{CE}=\dfrac{AB}{ED}\]
Replacing \[CA=CD+DA\] in the above equation, we get
\[\Rightarrow \dfrac{CD+DA}{CE}=\dfrac{AB}{ED}\]
Putting the values of each side by using the figure, we get
\[\Rightarrow \dfrac{8+7}{10}=\dfrac{9}{x}\]
Simplifying the above, we get
\[\Rightarrow \dfrac{15}{10}=\dfrac{9}{x}\]
Cross multiplied the above numerical, we obtain
\[\Rightarrow 15x=9\times 10\]
\[\Rightarrow 15x=90\]
Dividing both the side of the equation by 15, we get
\[\Rightarrow x=6\]
\[\therefore x=6cm\]

Note: We should always remember the property of congruence of two triangles and the similarity of two triangles. For the congruence of two given triangles sides and angles need to be equal but for similarity of two triangles, it is not necessary that angles be equal or sides are equal.