2. Proving Triangle Similarity by AA - 8. Similarity

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= DFAC, and using vertex A as the center of dilation. Let △ AB'C' be the image of this dilation.

Since dilations are similarity transformations, we know that the ratio of AC' to AC is equal to the scale factor. Recall that the scale factor was defined as DFAC. Therefore, by the Transitive Property of Equality, we have that AC'AC is equal to DFAC. AC'/AC = k DF/AC = k ⇒ AC'/AC = DF/AC Let's simplify the obtained equation.

AC'/AC = DF/AC

▼

Simplify

MultEqn

LHS * AC=RHS* AC

AC'/AC * AC= DF/AC * AC

CancelCommonFac

Cancel out common factors

AC'/AC * AC= DF/AC * AC

SimpQuotProd

Simplify quotient and product

AC'=DF

By definition of congruent sides, we have that AC'≅ DF. Next, we will pay close attention to the angles. Because corresponding angles of similar triangles are congruent, we know that ∠ C' ≅ ∠ C. We already knew that ∠ C ≅ ∠ F. By the Transitive Property of Congruence, we know that ∠ C' ≅ ∠ F. ∠ C' ≅ ∠ C ∠ C ≅ ∠ F ⇒ ∠ C' ≅ ∠ F Also, by the Reflexive Property of Congruence, we can say that ∠ A ≅ ∠ A. Next, let's draw △ AB'C' and △ DEF and list the congruent parts between them.

By the Angle-Side-Angle Congruence Theorem, we can state that △ AB'C' ≅ △ DEF. Therefore, there is a composition of rigid motions that maps △ AB'C' 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 ~ △ DEF.

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

Exercise 2 Page 427

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