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4.05 Corresponding parts of congruent triangles

Lesson

Concept summary

We have now covered all five theorems for triangle congruency and when they can be used:

  • Side-Side-Side (SSS) is when three sides of one triangle are congruent to the three sides of the other triangle.
  • Side-Angle-Side (SAS) is when both triangles have two congruent sides and the included angles are also congruent.
  • Angle-Side-Angle (ASA) is when both triangles have two congruent angles and the included sides are also congruent.
  • Angle-Angle-Side (AAS) is when both triangles have two congruent angles and a pair of corresponding non-included sides are congruent.
  • Hypotenuse-Leg (HL) is when both triangles have a right angle, congruent hypotenuses and a congruent leg.

Once we've established congruency between two triangles we can then justify the congruence of any pair of corresponding parts.

Corresponding parts of congruent triangles theorem

Corresponding parts of congruent triangles are congruent.

Two triangles. The 3 sides and 3 angles of one triangle are congruent to the other triangle.

We can reference this theorem with its acronym, CPCTC.

Worked examples

Example 1

Consider the triangles in the diagram.

Triangle A B C and E F D. Sides A B and E F are congruent, as well as sides A C and E D. Angles B A C and D E F are congruent.

Identify the angle that is congruent to \angle{ACB} and justify their congruence.

Approach

These angles are not labeled as congruent and are not any of the congruent angle pairs we recognize such as vertical angles, right angles, or corresponding angles. The only way we can justify their congruence is if the triangles themselves are congruent. First, justify the congruence of the triangles. Then justify the congruence of the angles by using the corresponding parts of congruent triangles theorem.

Solution

From the labels in the diagram, we can see that \overline{AB}\cong \overline{EF}, \angle{A}\cong \angle{E}, and \overline{AC}\cong \overline{ED}. We can conclude that \triangle{ABC} \cong \triangle{EFD} by the side-angle-side congruence theorem. Since \angle{ACB} corresponds to \angle{EDF}, we can justify that \angle{ACB}\cong \angle{EDF} because corresponding parts of congruent triangles are congruent.

Example 2

Jessah started the following explanation trying to show \triangle{RMP} is an isosceles triangle, but it is incomplete.

It is given in the diagram that \overline{RQ}\cong \overline{QP} and that \overline{RP}\perp \overline{MQ}. So \angle{MQR} and \angle{MQP} are right angles.

Complete the explanation showing \triangle{RMP} is an isosceles triangle.

Approach

Make sure the diagram is labelled clearly with the information that can be deduced by the information already given. This will help us determine which triangle congruency theorem we are using. Sometimes it helps to work backwards with what we're trying to show, so we know we need to show \triangle{RMP} is an isosceles, therefore we need to show that \overline{MR}\cong \overline{MP}. Since we weren't given any information about the length of these sides, the only way to show they're equal is to justify why \triangle RMQ \cong \triangle PMQ.

Here is an updated diagram, using the information already given on the diagram previously:

Triangle R M P with midpoint Q of side R P. A segment is drawn from M to Q perpendicular to R P, forming two right triangles.

From this labeling we can see that these triangles are congruent by side-angle-side. We're now ready to complete our explanation.

Solution

It is given in the diagram that \overline{RQ}\cong \overline{QP} and that \overline{RP}\perp \overline{MQ}. So \angle{MQR} and \angle{MQP} are right angles. Since \overline{MQ} is a shared side, we can see that \triangle RMQ \cong \triangle PMQ by Side-Angle-Side (SAS) congruence. So we know \overline{MR}\cong \overline{MP} since corresponding parts of congruent triangles are congruent (CPCTC). Therefore, \triangle{RMP} is an isosceles.

Example 3

Use the diagram to find the width of the arch measured from the outer edges of its pillars.

A diagram of an arch. On the bottom of the arch, an upside down triangle with a base length equal to the width of the arch is drawn. This triangle has leg lengths of 30 feet, and 26 feet. The apex of the upside down triangle intersects with the apex of another triangle. This triangle has leg lengths of 26 feet, and 30 feet, and base length of 24 feet. The 30 feet sides of the two triangles makes a straight segment. The 26 feet sides of the two triangles also makes a straight segment.

Approach

In order to find the length of the arch we need to justify the congruence of the triangles. Then we can find the length because it will be congruent to the side it corresponds to in the other triangle where all three lengths are known.

Solution

The sides of lengths 26\text{ ft} and 30\text{ ft} are congruent pairs because they have the same lengths, and the angles included by these sides are congruent because vertical angles are congruent. This means that the two triangles are congruent by side-angle-side congruence. The width of the arch corresponds to the side of the other triangle that is 24\text{ ft} long, and so the width of the arch is 24\text{ ft} as well since corresponding parts of congruent triangles are congruent.

Outcomes

G.CO.B.6

Use the definition of congruence in terms of rigid motions to show that two triangles are congruent if and only if corresponding pairs of sides and corresponding pairs of angles are congruent.

G.MP1

Make sense of problems and persevere in solving them.

G.MP3

Construct viable arguments and critique the reasoning of others.

G.MP5

Use appropriate tools strategically.

G.MP6

Attend to precision.

G.MP7

Look for and make use of structure.

G.MP8

Look for and express regularity in repeated reasoning.

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