Question

In figure, if $\angle CAB=\angle CED$ prove that $∆CAB~∆CED$

Hence, find the value of $x$

Answer 1

Step 1: Note the given data

Given the angles $\angle CAB=\angle CED$
The length of $CD=8units$
The length of $AD=7units$
The length of $CE=10units$
The length of $AB=9units$

AA similarity: If two angles of a triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

Applying AA theorem,

In $∆CABand∆CED$

$\angle CAB=\angle CED$………………………………………..….[Given]

$\angle ACB=\angle ECD$……………………………………………[Common angle]

$∆CAB~∆CED$……………………………….……………[by AA similarity]

Step 2: Calculate the value of $x$

Similar Triangles property: If two triangles are similar, then their corresponding sides are proportional.

Since $∆CAB~∆CED$, so

$\frac{CA}{CE}=\frac{AB}{ED}$

$\Rightarrow \frac{CD+AD}{CE}=\frac{AB}{ED}$

$\Rightarrow \frac{8+7}{10}=\frac{9}{x}$

$\Rightarrow x=\frac{9\times 10}{15}$

$\Rightarrow x=6units$

Hence, the value of $x$is $6units$.