Burhan Hopper

2021-01-28

To check: whether the additional information in the given option would be enough to prove the given similarity.
Given:
The given similarity is $\mathrm{△}ADC\sim \mathrm{△}BEC$

The given options are:
A.$\mathrm{\angle }DAC\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{\angle }ECB$ are congruent.
B.$\stackrel{―}{AC}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\stackrel{―}{BC}$ are congruent.
C.$\stackrel{―}{AD}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\stackrel{―}{EB}$ are parallel.
D.$\mathrm{\angle }CEB$ is a right triangle.

Tuthornt

Calculation:
The given similar triangle is $\mathrm{△}ADC\sim \mathrm{△}BEC$. And this is only possible $\stackrel{―}{AD}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\stackrel{―}{EB}$ are parallel.
Therefore, $\mathrm{\angle }CAD\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{\angle }CDA$ is equal to the $\mathrm{\angle }CBE\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{\angle }CEB$, so that the triangle is similar by AA-similarity
Hence, the correct option is (C).

Do you have a similar question?