Now that we can recognize similar triangles, we now demonstrate how to formally prove that two triangles are similar. These proofs will be a variation on the kinds of proofs we encountered when demonstrating congruence.
Before we get started, note that we write "the measures of two pairs of sides are in the same proportion" using fractions. So in a pair of triangles like these:
... if we are given $\frac{AB}{JL}=\frac{AC}{KL}=\frac{BC}{KJ}$ABJL=ACKL=BCKJ we know that the proportion of $AB$AB to $JL$JL is the same as the proportion of $AC$AC to $KL$KL, which is also the same as the proportion of $BC$BC to $KJ$KJ. This allows us to conclude that $\Delta ABC\sim\Delta JKL$ΔABC~ΔJKL, by $SSS\sim$SSS~.
Having established similarity, we can then solve for missing values, like we did for congruence.
In the following triangles, $\frac{QZ}{SZ}=\frac{PZ}{RZ}$QZSZ=PZRZ. Prove that these triangles are similar.
Think: We have been given two side measures in the same proportion, and the angles in between are vertical angles. We should use $SAS\sim$SAS~ to prove these triangles are similar.
Do: We write down what we want to prove, followed by the information obtained from the information given to us:
To prove: $\Delta PQZ\sim\Delta RSZ$ΔPQZ~ΔRSZ | |
Statement | Reason |
$\frac{QZ}{SZ}=\frac{PZ}{RZ}$QZSZ=PZRZ | Given |
$\angle PZQ\cong\angle RZS$∠PZQ≅∠RZS | Vertical angles are congruent |
$\Delta PQZ\sim\Delta RSZ$ΔPQZ~ΔRSZ | $SAS\sim$SAS~ |
Prove that $\Delta WXZ$ΔWXZ is similar to $\Delta XYZ$ΔXYZ, and solve for the value of $t$t:
Think: We have been given that there are two pairs of congruent angles, so we can demonstrate that these triangles are similar using AA. Once that is done, we can use the fact that these triangles are similar to solve for $t$t after matching up the sides correctly.
Do: We write down what we want to find, followed by the information obtained from the markings and the measures given on the diagram:
To find: the value of $t$t | |
Statement | Reason |
$\angle WZX\cong\angle XZY$∠WZX≅∠XZY | Given |
$\angle WXZ\cong\angle XYZ$∠WXZ≅∠XYZ | Given |
$\Delta WXZ\sim\Delta XYZ$ΔWXZ~ΔXYZ | $AA\sim$AA~ |
$ZW=1$ZW=1 | Given |
$ZX=3$ZX=3 | Given |
$ZY=t$ZY=t | Given |
$\frac{ZY}{ZX}=\frac{ZX}{ZW}$ZYZX=ZXZW | Corresponding sides of similar triangles are in the same proportion |
$\frac{t}{3}=\frac{3}{1}$t3=31 | Substitution |
$t=9$t=9 | Multiplication property of equality |
Prove that $\Delta CGD$ΔCGD and $\Delta EDF$ΔEDF are similar, and solve for the value of $\theta$θ:
Think: Here we are only given one angle measure that doesn't involve a variable, so we will need to rely on the sides to demonstrate similarity. Once that is done, we can match up the angles and form an equation to solve for $\theta$θ.
Do: We write down what we want to find, followed by the information obtained from the markings and the measures on the diagram:
To find: the value of $\theta$θ | |
Statement | Reason |
$CG=4$CG=4 | Given |
$ED=2$ED=2 | Given |
$\frac{CG}{ED}=\frac{4}{ED}$CGED=4ED | Multiplication property of equality |
$\frac{CG}{ED}=2$CGED=2 | Substitution |
$CD=5$CD=5 | Given |
$EF=2.5$EF=2.5 | Given |
$\frac{CD}{EF}=\frac{5}{EF}$CDEF=5EF | Multiplication property of equality |
$\frac{CD}{EF}=2$CDEF=2 | Substitution |
$GD=GF+FD$GD=GF+FD | Segment addition theorem |
$GF=FD$GF=FD | Given |
$GD=2FD$GD=2FD | Substitution |
$\frac{GD}{FD}=2$GDFD=2 | Multiplication property of equality |
$\frac{CG}{ED}=\frac{CD}{EF}=\frac{GD}{FD}$CGED=CDEF=GDFD | Transitive property of equality |
$\Delta CGD\sim\Delta EDF$ΔCGD~ΔEDF | $SSS\sim$SSS~ |
$\angle CGD=\angle EDF$∠CGD=∠EDF | Corresponding angles of similar triangles are congruent |
$m\angle CGD=m\angle EDF$m∠CGD=m∠EDF | Congruent angles have equal measure |
$m\angle CGD=54^\circ$m∠CGD=54° | Given |
$m\angle EDF=5\theta-6^\circ$m∠EDF=5θ−6° | Given |
$54=5\theta-6$54=5θ−6 | Substitution |
$60=5\theta$60=5θ | Addition property of equality |
$5\theta=60$5θ=60 | Symmetric property of equality |
$\theta=12$θ=12 | Multiplication property of equality |
This two-column proof shows that two triangles in the attached diagram are similar, but it is incomplete.
Statements | Reasons |
---|---|
$\angle BAC\cong\angle EDC$∠BAC≅∠EDC | Given |
$\angle BCA\cong\angle ECD$∠BCA≅∠ECD |
Vertical angles congruence theorem |
$\left[\text{____}\right]$[____] | $\left[\text{____}\right]$[____] |
Select the correct statement and reason to complete the proof.
$\Delta ABC\sim\Delta DEC$ΔABC~ΔDEC | Angle-angle similarity (AA$\sim$~) |
$\Delta ABC\sim\Delta DEC$ΔABC~ΔDEC | Side-angle-side similarity (SAS$\sim$~) |
$\Delta ABC\sim\Delta EDC$ΔABC~ΔEDC | Angle-angle similarity (AA$\sim$~) |
$\Delta ABC\sim\Delta DEC$ΔABC~ΔDEC | Side-side-side similarity (SSS $\sim$~) |
$\Delta ABC\sim\Delta EDC$ΔABC~ΔEDC | Side-side-side similarity (SSS $\sim$~) |
$\Delta ABC\sim\Delta EDC$ΔABC~ΔEDC | Side-angle-side similarity (SAS$\sim$~) |
$\Delta ABC\sim\Delta DEC$ΔABC~ΔDEC | Angle-angle similarity (AA$\sim$~) |
$\Delta ABC\sim\Delta DEC$ΔABC~ΔDEC | Side-angle-side similarity (SAS$\sim$~) |
$\Delta ABC\sim\Delta EDC$ΔABC~ΔEDC | Angle-angle similarity (AA$\sim$~) |
$\Delta ABC\sim\Delta DEC$ΔABC~ΔDEC | Side-side-side similarity (SSS $\sim$~) |
$\Delta ABC\sim\Delta EDC$ΔABC~ΔEDC | Side-side-side similarity (SSS $\sim$~) |
$\Delta ABC\sim\Delta EDC$ΔABC~ΔEDC | Side-angle-side similarity (SAS$\sim$~) |
This two-column proof shows that $\Delta ABC\sim\Delta XYZ$ΔABC~ΔXYZ in the attached diagram, but it is incomplete.
Statements | Reasons |
---|---|
$\frac{AC}{XZ}=\frac{BC}{ZY}$ACXZ=BCZY | Given |
$\angle ACB\cong\angle XZY$∠ACB≅∠XZY |
Given |
$\Delta ABC\sim\Delta XYZ$ΔABC~ΔXYZ |
$\left[\text{____}\right]$[____] |
Select the correct reason to complete the proof.
Angle-angle similarity (AA$\sim$~)
Side-angle-side similarity (SAS$\sim$~)
Side-side-side similarity (SSS$\sim$~)
Angle-angle similarity (AA$\sim$~)
Side-angle-side similarity (SAS$\sim$~)
Side-side-side similarity (SSS$\sim$~)
This two-column proof shows that two triangles in the attached diagram are similar, but it is incomplete.
Statements | Reasons |
---|---|
$\frac{AC}{YZ}=\frac{AB}{XZ}=\frac{CB}{XY}$ACYZ=ABXZ=CBXY | Given |
$\left[\text{____}\right]$[____] | $\left[\text{____}\right]$[____] |
Select the correct statement and reason to complete the proof.
$\Delta ABC\sim\Delta XYZ$ΔABC~ΔXYZ | Side-side-side similarity (SSS$\sim$~) |
$\Delta ABC\sim\Delta ZYX$ΔABC~ΔZYX | Side-angle-side similarity (SAS $\sim$~) |
$\Delta ABC\sim\Delta ZXY$ΔABC~ΔZXY | Side-side-side similarity (SSS$\sim$~) |
$\Delta ABC\sim\Delta ZXY$ΔABC~ΔZXY | Side-angle-side similarity (SAS$\sim$~) |
$\Delta ABC\sim\Delta XYZ$ΔABC~ΔXYZ | Side-side-side similarity (SSS$\sim$~) |
$\Delta ABC\sim\Delta ZYX$ΔABC~ΔZYX | Side-angle-side similarity (SAS $\sim$~) |
$\Delta ABC\sim\Delta ZXY$ΔABC~ΔZXY | Side-side-side similarity (SSS$\sim$~) |
$\Delta ABC\sim\Delta ZXY$ΔABC~ΔZXY | Side-angle-side similarity (SAS$\sim$~) |
Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures.