91Ó°ÊÓ

Understanding the relationship between parallel lines can unlock the secrets of angle measurements in geometry. Imagine two train tracks running side by side. No matter how far they go, they never meet—this is what we mean by parallel. When a third line crosses these tracks, it creates several angles.

In our exercise, lines \( \overline{F G} \) and \( \overline{J K} \) are parallel. When a line such as \( \overline{G H} \), known as a transversal, crosses these parallel lines, alternate interior angles are formed. Alternate interior angles are on opposite sides of the transversal and inside the parallel lines. A neat trick here is that these angles are always equal.

Let's say in our case, angle \( FGH \) and angle \( JHJ \) are alternate interior angles. Since \( \overline{F G} \parallel \overline{J K} \), we know \( \angle FGH = \angle JHJ \). This relationship is crucial for figuring out the exercises involving parallel lines and the angles they create.

A 45-45-90 triangle is a very special type of right triangle. Named for its angles, it features two angles of \(45^{\circ}\) and one right angle of \(90^{\circ}\). This triangle is also known as an isosceles right triangle. Why? Because its two legs are always of equal length.

What makes 45-45-90 triangles useful is their consistent ratio of side lengths. For a triangle with legs of length \(a\):

• Each leg is equal to \(a\).
• The hypotenuse is \(a\sqrt{2}\).

In the exercise, triangles \(G HF\) and \(F HK\) form 45-45-90 triangles. Knowing this, their construction split into two equal \(45^{\circ}\) angles made solving for other angles seamless. Here, angle \(F\) equals angle \(H\), simplifying the calculations.

The Exterior Angle Theorem offers a handy tool when solving angle problems involving triangles. This theorem tells us something important: the measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles.

Picture a triangle where you "extend" one side to form an exterior angle. Now, take the two opposite inside angles -- their sum gives you the measure of the exterior angle.

In our problem, consider triangle \(JHK\) with exterior angle \(H\). According to the theorem:

• The exterior angle \( \angle H \) equals \( \angle JHK + \angle F \).

By applying the theorem, and knowing both \( \angle JHK \) and \( \angle F \) are \(45^{\circ}\), we accurately determined \(\angle J\) to be \(90^{\circ}\), making problem solving much easier!