Proof:

Consider the given triangle,

The \(\displaystyle\angle{A}\) is part of all three triangles \(\displaystyle\triangle{A}{B}{C},\triangle{A}{D}{E}{\quad\text{and}\quad}\triangle{A}{F}{G}\) where, \(\displaystyle\angle{C}=\angle{E}=\angle{G}={90}^{{\circ}}\).

Since , sum of all angles of a triangle is \(\displaystyle{180}^{{\circ}}\) so from the triangle ABC,

\(\displaystyle\angle{A}+\angle{B}+\angle{C}={180}^{{\circ}}\)

\(\displaystyle\angle{B}={180}^{{\circ}}-\angle{A}-\angle{C}\)

\(\displaystyle={180}^{{\circ}}-\angle{A}-{90}^{{\circ}}\)

\(\displaystyle={90}^{{\circ}}-\angle{A}\)

In the triangle \(\displaystyle\triangle{A}{D}{E}\),

\(\displaystyle\angle{A}+\angle{D}+\angle{E}={180}^{{\circ}}\)

\(\displaystyle\angle{D}={180}^{{\circ}}-\angle{A}-\angle{E}\)

\(\displaystyle={180}^{{\circ}}-\angle{A}-{90}^{{\circ}}\)

\(\displaystyle={90}^{{\circ}}-\angle{A}\)

Similarly, from triangle AFG the angle \(\displaystyle\angle{F}={90}^{{\circ}}—{Z}{A}\).

Therefore, \(\displaystyle\angle{B}=\angle{D}=\angle{F}\).

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.

Consider the given triangle,

The \(\displaystyle\angle{A}\) is part of all three triangles \(\displaystyle\triangle{A}{B}{C},\triangle{A}{D}{E}{\quad\text{and}\quad}\triangle{A}{F}{G}\) where, \(\displaystyle\angle{C}=\angle{E}=\angle{G}={90}^{{\circ}}\).

Since , sum of all angles of a triangle is \(\displaystyle{180}^{{\circ}}\) so from the triangle ABC,

\(\displaystyle\angle{A}+\angle{B}+\angle{C}={180}^{{\circ}}\)

\(\displaystyle\angle{B}={180}^{{\circ}}-\angle{A}-\angle{C}\)

\(\displaystyle={180}^{{\circ}}-\angle{A}-{90}^{{\circ}}\)

\(\displaystyle={90}^{{\circ}}-\angle{A}\)

In the triangle \(\displaystyle\triangle{A}{D}{E}\),

\(\displaystyle\angle{A}+\angle{D}+\angle{E}={180}^{{\circ}}\)

\(\displaystyle\angle{D}={180}^{{\circ}}-\angle{A}-\angle{E}\)

\(\displaystyle={180}^{{\circ}}-\angle{A}-{90}^{{\circ}}\)

\(\displaystyle={90}^{{\circ}}-\angle{A}\)

Similarly, from triangle AFG the angle \(\displaystyle\angle{F}={90}^{{\circ}}—{Z}{A}\).

Therefore, \(\displaystyle\angle{B}=\angle{D}=\angle{F}\).

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.