Exercise 2 Page 427

Let's consider two triangles with two pairs of corresponding congruent angles.

We want to establish a relation between the triangles above. To do so, we will start by performing a dilation on $△ABC$ by a scale factor of $k=\frac{DF}{AC},$ and using vertex $A$ as the center of dilation. Let $△A{B}^{\prime }{C}^{\prime }$ be the image of this dilation.

Since dilations are similarity transformations, we know that the ratio of $A{C}^{\prime }$ to $AC$ is equal to the scale factor. Recall that the scale factor was defined as $\frac{DF}{AC}.$ Therefore, by the Transitive Property of Equality, we have that $\frac{A{C}^{\prime }}{AC}$ is equal to $\frac{DF}{AC}.$ Let's simplify the obtained equation. By definition of congruent sides, we have that $\overline{A{C}^{\prime }}\cong \overline{DF}.$ Next, we will pay close attention to the angles. Because corresponding angles of similar triangles are congruent, we know that $\angle C^{\prime }\cong \angle C.$ We already knew that $\angle C\cong \angle F.$ By the Transitive Property of Congruence, we know that $\angle C^{\prime }\cong \angle F.$ Also, by the Reflexive Property of Congruence, we can say that $\angle A\cong \angle A.$ Next, let's draw $△A{B}^{\prime }{C}^{\prime }$ and $△DEF$ and list the congruent parts between them.

By the Angle-Side-Angle Congruence Theorem, we can state that $△A{B}^{\prime }{C}^{\prime }\cong △DEF.$ Therefore, there is a composition of rigid motions that maps $△A{B}^{\prime }{C}^{\prime }$ to $△DEF.$

This means that we can map $△ABC$ onto $△DEF$ by performing a dilation followed by a composition of rigid motions. Consequently, we can map $△ABC$ to $△DEF$ by applying a similarity transformation, which means that $△ABC\sim △DEF.$

If two triangles have two pairs of corresponding angles that are congruent, then the triangles are similar.