Exercise 3 Page 435

We want to find two different ways to prove that two triangles are similar by using their corresponding sides. We will consider two cases.

1. The triangles have three corresponding proportional sides.
2. The triangles have two pairs of corresponding proportional sides and congruent included angles.

Let's do it!

Three Corresponding Proportional Sides

Let △ ABC and △ PQR be two triangles with three corresponding proportional sides.

We want to know if the information above is enough to prove the triangles are similar. Let Y be a point on 1PR such that PY= AC. Then, we will draw XY so that XY∥ QR.

By the Corresponding Angles Theorem, we know that ∠ PYX ≅ ∠ PRQ. Moreover, ∠ P, which is common for △ PXY and △ PQR, is congruent to itself by the Reflexive Property of Congruence. Therefore, △ PXY and △ PQR are similar triangles by the Angle-Angle Similarity Theorem.

Consequently, we can write the following proportions. PQ/PX = PR/PY = QR/XY Let's use the fact that PY= AC and substitute it in the equation PQ PX = PR PY. PQ/PX = PR/PY PY= AC ⟶ PQ/PX = PR/AC Now, we will substitute PQ AB for PR AC, and simplify.

From the above, we have that AB ≅ PX.

Next, since PY= AC, we can substitute it in the equation PR PY = QRXY. PR/PY = QR/XY PY= AC ⟶ PR/AC = QR/XY Now, will use the fact that PR AC= QR BC and substitute it into the equation above.

Therefore, XY≅ BC.

By the Side-Side-Side Congruence Theorem, we have that △ ABC ≅ △ PXY. With this, and knowing that △ PXY ~ △ PQR, we can say that △ ABC ~ △ PQR. Note that we started supposing that △ ABC and △ PQR had three pairs of corresponding proportional sides.

Two Pairs of Corresponding Proportional Sides and Congruent Included Angles

Let △ ABC and △ PQR be two triangles with two corresponding proportional sides and congruent included angles.

As we did before, let Y be a point on PR such that PY= AC. Then, we will draw XY so that XY∥ QR.

By the Corresponding Angles Theorem, we have that ∠ PYX ≅ ∠ PRQ. Furthermore, ∠ P, which is common for △ PXY and △ PQR, is congruent to itself by the Reflexive Property of Congruence. Therefore, △ PXY and △ PQR are similar triangles by the Angle-Angle Similarity Theorem. PQ/PX = PR/PY = QR/XY Let's substitute PY = AC in the equation PQ PX = PR PY. PQ/PX = PR/PY PY= AC ⟶ PQ/PX = PR/AC Let's now substitute PQ AB for PR AC into the equation above, and simplify.

The latter equation implies that AB ≅ PX.

By the Side-Angle-Side Congruence Theorem, we know that △ ABC ≅ △ PXY. Since corresponding parts of congruent figures are congruent, we have that ∠ ACB ≅ ∠ PYX. Finally, since ∠ PYX ≅ ∠ PRQ, we obtain that ∠ ACB ≅ ∠ PRQ by the Transitive Property of Congruence. ∠ ACB ≅ ∠ PYX ∠ PYX ≅ ∠ PRQ ⇓ ∠ ACB ≅ ∠ PRQ By the Angle-Angle Similarity Theorem, we conclude that △ ABC ~ △ PQR, which is what we wanted to show.