Unit 4, Day 4: Congruence Proofs

Congruence Proofs

Using Algebra & the Coordinate Plane

Today's Objective

To use algebra and coordinate geometry (the distance formula) to prove that two triangles are congruent.

Do Now: Review

1. Solve for x: \(3x - 5 = 16\)

1. Solve for x: \(3x - 5 = 16\)

2. What are the three congruence postulates we learned yesterday?

2. What are the three congruence postulates we learned yesterday?

Congruence with Algebra

If we know two triangles are congruent, we can set up and solve equations because their corresponding parts must be equal.

CPCTC: Corresponding Parts of Congruent Triangles are Congruent.

How to Solve Algebraically

Step 1: Read the congruence statement (e.g., \( \triangle ABC \cong \triangle DEF \)) to identify which parts correspond.

Step 2: Find the corresponding parts that have algebraic expressions.

Step 3: Set the expressions for the corresponding parts equal to each other.

Step 4: Solve the equation for the variable.

Practice: Solving for a Variable

I Do

Given \( \triangle JKL \cong \triangle MNO \), \( JK = 2x-4 \), \( MN = 12 \).

\( 2x - 4 = 12 \)

\( 2x = 16 \)

\( x = 8 \)

We Do

Given \( \triangle ABC \cong \triangle XYZ \), \( m\angle B = (3y+1)^\circ \), \( m\angle Y = 40^\circ \).

\( 3y + 1 = 40 \)

\( 3y = 39 \)

\( y = 13 \)

You Do

Given \( \triangle PQR \cong \triangle STU \), \( QR = z+7 \), \( TU = 2z \).

\( z+7 = 2z \)

\( z = 7 \)

I Do: Given \( \triangle JKL \cong \triangle MNO \), \( JK = 2x-4 \), \( MN = 12 \). Solve for x. x=8

We Do: Given \( \triangle ABC \cong \triangle XYZ \), \( m\angle B = (3y+1)^\circ \), \( m\angle Y = 40^\circ \). Solve for y.

You Do: Given \( \triangle PQR \cong \triangle STU \), \( QR = z+7 \), \( TU = 2z \). Solve for z.

Congruence on the Coordinate Plane

We can prove triangles are congruent by finding the lengths of their sides using the Distance Formula.

\( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

If all three pairs of sides are equal in length, the triangles are congruent by SSS.

How to Prove on a Coordinate Plane

Step 1: List the coordinates for all 6 vertices of the two triangles.

Step 2: Use the Distance Formula to calculate the length of all 6 sides.

Step 3: Match up the corresponding sides and compare their lengths.

Step 4: If all three pairs of sides are congruent, state that the triangles are congruent by SSS.

Practice: Coordinate Proofs

I Do

A(1,5), B(1,1), C(4,1)

AB=4, BC=3, AC=5

D(-1,-4), E(-1,0), F(-4,0)

DE=4, EF=3, DF=5

Congruent by SSS

We Do

P(2,1), Q(5,1), R(2,5)

S(-1,1), T(-4,1), U(-1,5)

PQ=3, QR=4, PR=5

ST=3, TU=5, SU=4

Wait... QR!=SU. Not congruent!

Let's try S(-1,1), T(-4,1), U(-1,-3). Then ST=3, TU=4, SU=5. Now they are congruent by SSS.

You Do

G(-2,-2), H(1,-2), I(-2,2)

J(3,3), K(6,3), L(3,7)

GH=3, HI=5, GI=4

JK=3, KL=4, JL=5

JK=GH, but KL!=HI. Not congruent!

For "We Do" and "You Do", let's fix the points to make them congruent and then prove it.

I Do: Prove \( \triangle ABC \cong \triangle DEF \). A(1,5), B(1,1), C(4,1) and D(-1,-4), E(-1,0), F(-4,0).
AB=4, BC=3, AC=5. DE=4, EF=3, DF=5. Congruent by SSS.

We Do: Prove \( \triangle PQR \cong \triangle STU \). P(2,1), Q(5,1), R(2,5) and S(-1,1), T(-4,1), U(-1,-3).

You Do: Prove \( \triangle GHI \cong \triangle JKL \). G(-2,-2), H(1,-2), I(-2,2) and J(3,3), K(6,3), L(3,7).

Independent Practice

Level 1 Problem:

1. Given \( \triangle CAT \cong \triangle DOG \), \( CA = 10 \), and \( DO = 3x-2 \). Find x.

Level 2 Problem:

2. Given \( \triangle LMN \cong \triangle OPQ \), \( m\angle M = 55^\circ \), and \( m\angle P = (4y+15)^\circ \). Find y.

Level 3 Problem:

3. Prove or disprove that \( \triangle ABC \cong \triangle XYZ \) using the distance formula. A(0,0), B(3,0), C(0,4) and X(1,1), Y(4,1), Z(1,5).

Independent Practice

1. Given \( \triangle CAT \cong \triangle DOG \), \( CA = 10 \), and \( DO = 3x-2 \). Find x.

2. Given \( \triangle LMN \cong \triangle OPQ \), \( m\angle M = 55^\circ \), and \( m\angle P = (4y+15)^\circ \). Find y.

3. Prove or disprove that \( \triangle ABC \cong \triangle XYZ \) given A(0,0), B(3,0), C(0,4) and X(1,1), Y(4,1), Z(1,5).

Exit Ticket

Given \( \triangle WXY \cong \triangle R S T \), \( WX = 15 \), \( XY = 20 \), \( WY = 25 \), and \( ST = 3z - 1 \). Find the value of z.

\( XY \) corresponds to \( ST \).

\( 20 = 3z - 1 \)

\( 21 = 3z \)

z = 7

Answer: