FizeauV

2021-02-02

Proving Similarity in the figure DEFG is a square. Prove the following:
$\mathrm{△}ADG\sim \mathrm{△}GCF$
Given:
The given figure is,

Maciej Morrow

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.
Calculation:
It is given that DEFG is a square.
Consider $\mathrm{△}ADG\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{△}GCF$.
$\mathrm{\angle }ADG=\mathrm{\angle }GCF$ both are ${90}^{\circ }$.
And $GF\mid \mid AB$, thus, $\mathrm{\angle }CGF=\mathrm{\angle }GAD$ as they are alternate angles.
Therefore, $\mathrm{△}ADG\sim \mathrm{△}GCF$ by AA rule.
Conclusion:
Hence, it is proved that $\mathrm{△}ADG\sim \mathrm{△}GCF$.

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