 # Similar Triangles and Triginimetric Functions Use the figure below. Explain why /_ABC. /_ADE, and /_AFG are similar triangles. avissidep 2021-02-06 Answered
Similar Triangles and Triginimetric Functions Use the figure below.
Explain why $\mathrm{△}ABC.\mathrm{△}ADE,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{△}AFG$ are similar triangles.
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Proof:
Consider the given triangle, The $\mathrm{\angle }A$ is part of all three triangles $\mathrm{△}ABC,\mathrm{△}ADE\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{△}AFG$ where, $\mathrm{\angle }C=\mathrm{\angle }E=\mathrm{\angle }G={90}^{\circ }$.
Since , sum of all angles of a triangle is ${180}^{\circ }$ so from the triangle ABC,
$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$
$\mathrm{\angle }B={180}^{\circ }-\mathrm{\angle }A-\mathrm{\angle }C$
$={180}^{\circ }-\mathrm{\angle }A-{90}^{\circ }$
$={90}^{\circ }-\mathrm{\angle }A$
In the triangle $\mathrm{△}ADE$,
$\mathrm{\angle }A+\mathrm{\angle }D+\mathrm{\angle }E={180}^{\circ }$
$\mathrm{\angle }D={180}^{\circ }-\mathrm{\angle }A-\mathrm{\angle }E$
$={180}^{\circ }-\mathrm{\angle }A-{90}^{\circ }$
$={90}^{\circ }-\mathrm{\angle }A$
Similarly, from triangle AFG the angle $\mathrm{\angle }F={90}^{\circ }—ZA$.
Therefore, $\mathrm{\angle }B=\mathrm{\angle }D=\mathrm{\angle }F$.
Since two of the angles of the triangles are equal, the third angle is also equal.
Therefore, the triangles are similar due to AAA similarity criterion.