# Proving Similarity In the figure CDEF is a rectangle. Prove that /_ ABC ~ /_EBF.Given:The given figure is,12210202681.jpg

Proving Similarity In the figure CDEF is a rectangle. Prove that $\mathrm{△}ABC\sim \mathrm{△}EBF$.
Given:
The given figure is,

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Pohanginah
Approach:
Two triangles are similar if their vertices can be matched up so that corresponding angles are congruent. In this case corresponding sides are proportional.
If the two angles of the triangles are same then the third angle of the triangles have to be same, because the sum of angles in a triangle is ${180}^{\circ }$. Therefore, the triangles are similar by AA rule if two angles are same.
Proof:
It is given that CDEF is a rectangle.
So, $\mathrm{\angle }C=\mathrm{\angle }EFB$.
$\mathrm{\angle }B$ is common in $\mathrm{△}ABC\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{△}EBF$.
Therefore, the two triangles are similar by AA rule.
Thus, $\mathrm{△}ABC\sim \mathrm{△}EBF$.
Similarly, $\mathrm{\angle }C=\mathrm{\angle }ADE$.
$\mathrm{\angle }A$ is common in $\mathrm{△}ABC\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{△}AED$.
Therefore, the two triangles are similar by AA rule.
Thus, $\mathrm{△}ABC\sim \mathrm{△}AED$.
Since, $\mathrm{△}ABC\sim \mathrm{△}EBF\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{△}ABC\sim \mathrm{△}AED,Thus,\mathrm{△}EBF\sim \mathrm{△}AED$.
Therefore, $\mathrm{△}ABC\sim \mathrm{△}AED\sim \mathrm{△}EBF$.
Conclusion:
Hence, it is proved that $\mathrm{△}ABC\sim \mathrm{△}AED\sim \mathrm{△}EBF$.