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The lines $CE$, $BF$, $AG$ are parallel because they have a common perpendicular line. By the intercept theorem, the triangles are indead similar and the scaling factor is 3, hence both height and base line of the bigger triangle are 3 times the corresponding length of the smaller, thus finally the area is 9 ties bigger.
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Yes, the triangles are similar. How can you prove this? (one way would be to find 2 sets of congruent angles)
Now that we know the triangles are similar, let's try and find the ratio of similarity. We could try and compare the bases of the triangles, the altitudes (the "right" sides) or the hypotenuses (the "left" sides). Which do we know the most about? (Hint: if AB = BC = CD, what does that say about BD and AD?)
Finally, yes, there is a property regarding similar triangles and their areas. See if you can deduce what it is - in this case you know how big the base and height of each triangle are in proportion to each other, so it should be easy to figure this out.
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Because of similar triangles we have that $$ \frac{DE}{DG}=\frac{CE}{AG}=\frac{1}{3}\longrightarrow \frac{DE}{DG}\cdot\frac{CE}{AG}=\frac{1}{9}\longrightarrow\frac{\displaystyle\frac{ DE \cdot CE}{2}}{\displaystyle\frac{DG\cdot AG}{2}}=\frac{42}{\triangle ADG}=\frac{1}9\longrightarrow \triangle ADG=378.$$
Q.E.D.
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The right triangles CDE and ADG are similar since there is the common angle at D, the right angle and third angle (determined since triangles have 180 degrees). So all sides are in the same proportion. Since $AD=3CD$, then the other sides, base $AG=3CE$, and height $GD=3ED$. Using the formula for the area as $\frac{1}{2}$base$\cdot$ height, the area of the larger triangle is 9 times the smaller area or $9\cdot 42=378$.
As $CE||BF||AG,$
So, $\frac{DC}{DE}=\frac{DB}{DF}=\frac{DA}{DG}$
But $DA=DC+CB+BA=3DC\implies DG=3DE$ and $AG=3CE$
Now $\triangle CDE=\frac{1}{2}\cdot EC \cdot DE=42$(given),
$\triangle ADG =\frac{1}{2}\cdot AG\cdot DG=\frac{1}{2}\cdot3CE\cdot 3DE=9\cdot \frac{1}{2}\cdot CE \cdot DE=9\cdot 42=378$