Given
that [tex]|A C|[/tex] and [tex]|A E|[/tex] are common external tangents of circle G and circle D.
We recall that the lengths of the tangents to a circle drawn from the same point are equal.
i.e. [tex]|A F| = |A B|[/tex] and [tex]|A E| = |A C|[/tex]
Thus, we find the length of [tex]|A E|[/tex] which is equal to the length of [tex]|A C|[/tex].
But [tex]|A E| = |A F| + |F E|[/tex].
Given that [tex]|F E| = 15[/tex], to find [tex]|A F|[/tex] we recall that the tangent of any given circle makes a right angle with the radius (or diameter) of the circle at the point of tangency.
We then use the pythagora rule to find the length [tex]|A F|[/tex].
Notice that triangle AFG is a right triangle with the side [tex]|A G|[/tex] being the hypothenus.
Recall that the pythagoras theorem states that the square of the hypothenus is equal to the sum of the square of the lengths of the other two legs of a right triangle.
Thus,
[tex]|A G|^2=|A F|^2+|G F|^2 \\ \\ |A F|^2=|A G|^2-|G F|^2 \\ \\ |A F|^2=5^2-3^2=25-9=16 \\ \\ |A F|= \sqrt{16} =4[/tex]
Thus, [tex]|A E| = 4 + 15 = 19[/tex] and hence [tex]|A C| = 19[/tex].
