Exercise 25 Page 275

Let's illustrate the toy and name the angles. At the same time, we note that the two triangles share a side, BC. By the Reflexive Property of Congruence, we know that these sides are congruent. Let's add this information to the diagram.

Next, we will replace the angle names with the expressions of the angles.

From the congruence statement △ A B C ≅ △ D B C, we know that ∠ B CA and ∠ B CD are corresponding. Similarly, we know that ∠ A BC and ∠ D BC are corresponding. With this, we can write two equations. 5x+10^(∘) &=3y+2^(∘) 8x-32^(∘) &=4y-24^(∘) Combining these equations, we get a system of equations which we can solve.

5x+10^(∘) =3y+2^(∘) & (I) 8x-32^(∘) =4y-24^(∘) & (II)

▼

(II): Solve for y

AddEqn

(II): LHS+24^(∘)=RHS+24^(∘)

5x+10^(∘) =3y+2^(∘) 8x-8^(∘) =4y

DivEqn

(II): .LHS /4^(∘).=.RHS /4^(∘).

5x=3y-8^(∘) 2x^(∘)-2^(∘) =y

RearrangeEqn

(II): Rearrange equation

5x^(∘)=3y-8^(∘) y=2x^(∘)-2^(∘)

Now that we have y equal to an expression, we can replace y with this expression in the first equation.

5x^(∘)=3y-8^(∘) & (I) y=2x^(∘)-2^(∘) & (II)

Substitute

(I): y= 2x^(∘)-2^(∘)

5x^(∘)=3( 2x^(∘)-2^(∘))-8^(∘) y=2x^(∘)-2^(∘)

▼

(I): Solve for x

Distr

(I): Distribute 3

5x^(∘)=6x^(∘)-6^(∘)-8^(∘) y=2x^(∘)-2^(∘)

SubTerm

(I): Subtract term

5x^(∘)=6x^(∘)-14^(∘) y=2x^(∘)-2^(∘)

SubEqn

(I): LHS-6x=RHS-6x

- x^(∘)=- 14^(∘) y=2x^(∘)-2^(∘)

MultEqn

(I): LHS * (- 1)=RHS* (- 1)

x^(∘)=14^(∘) y=2x^(∘)-2^(∘)

Having solved for x in Equation (I), we can solve for y by substituting this value in Equation (II).

x^(∘)=14^(∘) y=2x^(∘)-2^(∘)

Substitute

(II): x= 14^(∘)

x^(∘)=14^(∘) y=2( 14^(∘))-2^(∘)

▼

(II): Solve for y

Multiply

(II): Multiply

x^(∘)=14^(∘) y=28^(∘)-2^(∘)

SubTerm

(II): Subtract term

x^(∘)=14^(∘) y=26^(∘)

Having calculated both x and y, we will substitute x= 14^(∘) and y= 26^(∘) into the expressions for the two angle pairs. m∠ BCA:& 8( 14^(∘))-32^(∘)= 80^(∘) m∠ ABC:& 5( 14^(∘))+10^(∘)= 80^(∘) m∠ BCD:& 4( 26^(∘))-24^(∘)= 80^(∘) m∠ DBC:& 3( 26^(∘))+2^(∘) = 80^(∘) The base angles of the two triangles are equal and therefore congruent.

With this information, we can claim congruence by the ASA Congruence Theorem because we have two sets of congruent corresponding angles where the included sides are also congruent.