Question

state if the triangles in each pair are similar. if so, state how you know they are similar and complete the similarity statement.
1)

△pqr ~ ______
2)

△cba ~ ______

Explanation:

Problem 1:

Step 1: Identify Vertical Angles

$\angle QPR$ and $\angle EPF$ are vertical angles, so $\angle QPR = \angle EPF$.

Step 2: Check Proportional Sides

Calculate ratios of corresponding sides:
$\frac{QP}{FP} = \frac{20}{5} = 4$, $\frac{RP}{EP} = \frac{32}{8} = 4$.
Since $\frac{QP}{FP} = \frac{RP}{EP}$ and included angles are equal, by SAS similarity, $\triangle PQR \sim \triangle PEF$.

Step 1: Calculate Ratios of Sides

For $\triangle CBA$ and $\triangle CLK$:
$\frac{CB}{CK} = \frac{168}{60} = \frac{14}{5}$, $\frac{CA}{CL} = \frac{112}{72} = \frac{14}{9}$? Wait, no—wait, $CB = 168$, $CK = 60$? Wait, no, the side $CB$ is $168$, $CK$ is $60$? Wait, no, the segment from $C$ to $K$ is $60$, and from $K$ to $B$? Wait, no, the left side: total length $168$, and $CK = 60$, $KB = 168 - 60 = 108$? Wait, no, the triangle sides: $CB = 168$, $CK = 60$? Wait, no, the triangle $\triangle CBA$ has $CB = 168$, $BA = 196$, $CA = 112$. $\triangle CLK$ has $CK = 60$, $LK = 70$, $CL = 72$? Wait, no, $CL$ is $72$? Wait, $CA = 112$, $LA = 72$, so $CL = 112 - 72 = 40$? Wait, I misread. Let's correct:

$CA = 112$, $LA = 72$, so $CL = 112 - 72 = 40$? No, the diagram: $C$ to $L$ to $A$, with $CL$? Wait, the bottom side: $C$ to $L$ is? Wait, the labels: $C$, $L$, $A$, with $L$ to $A$ is $72$, $C$ to $A$ is $112$, so $C$ to $L$ is $112 - 72 = 40$? Wait, no, the problem says: $C$ to $L$ is? Wait, the triangle $\triangle CBA$: $CB = 168$, $BA = 196$, $CA = 112$. $\triangle CLK$: $CK = 60$, $LK = 70$, $CL = 40$? Wait, no, the left side: $CK = 60$, $KB = 168 - 60 = 108$? No, let's recalculate ratios properly.

Wait, $\frac{CK}{CB} = \frac{60}{168} = \frac{5}{14}$, $\frac{CL}{CA} = \frac{40}{112} = \frac{5}{14}$, $\frac{LK}{BA} = \frac{70}{196} = \frac{5}{14}$. Wait, no, $CL$: if $CA = 112$ and $LA = 72$, then $CL = 112 - 72 = 40$. So $\frac{CL}{CA} = \frac{40}{112} = \frac{5}{14}$, $\frac{CK}{CB} = \frac{60}{168} = \frac{5}{14}$, $\frac{LK}{BA} = \frac{70}{196} = \frac{5}{14}$. So all ratios are equal, so by SSS similarity, $\triangle CBA \sim \triangle CLK$. Wait, no: $CK = 60$, $CB = 168$? Wait, no, maybe $CB$ is $60 + KB$, but the left side is labeled $168$ as total? Wait, the left side: from $C$ up to the top is $168$, with $CK = 60$, so $KB = 168 - 60 = 108$. Then $CB = 168$, $CK = 60$, so $\frac{CK}{CB} = \frac{60}{168} = \frac{5}{14}$. $CA = 112$, $CL = 112 - 72 = 40$? No, $CL$ is $72$? Wait, the diagram: $C$ to $L$ is $72$? No, the bottom: $C$ to $L$ to $A$, with $L$ to $A$ is $72$, $C$ to $A$ is $112$, so $C$ to $L$ is $112 - 72 = 40$? I'm confused. Let's use the given numbers:

$\triangle CBA$: sides $CB = 168$, $BA = 196$, $CA = 112$.
$\triangle CLK$: sides $CK = 60$, $LK = 70$, $CL = 72$? Wait, no, $CL$ is $72$? Then $CA = 112$, $CL = 72$, so $LA = 112 - 72 = 40$? No, the problem says $L$ to $A$ is $72$. So $CL = 112 - 72 = 40$. Then $\frac{CL}{CA} = \frac{40}{112} = \frac{5}{14}$, $\frac{CK}{CB} = \frac{60}{168} = \frac{5}{14}$, $\frac{LK}{BA} = \frac{70}{196} = \frac{5}{14}$. So all three ratios are equal, so by SSS similarity, $\triangle CBA \sim \triangle CLK$. Wait, but $CL = 40$, $CK = 60$, $LK = 70$; $CA = 112$, $CB = 168$, $BA = 196$. Let's check ratios:

$\frac{CK}{CB} = \frac{60}{168} = \frac{5}{14}$, $\frac{CL}{CA} = \frac{40}{112} = \frac{5}{14}$, $\frac{LK}{BA} = \frac{70}{196} = \frac{5}{14}$. Yes, all equal. So $\triangle CBA \sim \triangle CLK$.

Answer:

$\triangle PEF$

Problem 2: