Exercise 18 Page 282

1. Assume that the points marked $A$ and $B$ below are midpoints of the square's sides.
2. Call the length of the square's side $2x.$
3. Assume that the yellow rectangle is a square.

Let's add this information to our diagram.

Next, we will draw a few segments that will help us decide which triangle could have an area that is twice that of the purple triangle. At the same time, we will introduce a few labels.

We can rule out the orange, green and brown triangles. That leaves us with the red triangle. If we can show that (2) and (3) are congruent with (1), we can prove that the purple triangle is twice as large as the red triangle.

Triangle (1) and (2)

We have already been given one pair of congruent sides. Additionally, there is a second pair of congruent sides, the ones with a side length of $x.$ Finally, we can identify two vertical angles at $B.$ According to the Vertical Angles Congruence Theorem, vertical angles are congruent. Let's add this information to our diagram.

By the SAS Congruence Theorem we can prove that $(1)$ and $(2)$ are congruent which means they have the same area.

Triangle (2) and (3)

There are a few things we can say about $(2)$ and $(3).$ First, they share a side so by the Reflexive Property of Congruence, they are congruent. Additionally, the sides with length $x$ will be congruent as well. Let's add all of this information to the diagram.

Finally, we can mark a couple of right angles, one because it forms a linear pair with one of the yellow squares angles, and another because it's a vertical angle to the same angle.

Since both (2) and (3) are right triangles with one pair of congruent legs and congruent hypotenuses, we can prove congruence by the HL Congruence Theorem. Since $(1)$ is congruent to both $(2)$ and $(3),$ it must be that the area of the red triangle is twice that of the purple triangle.