Question

practice describing supplementary angle relationships
study the example showing how to use angle relationships to solve problems. then solve problems 1–8.
example
what is the value of x?
∠acd and ∠hfg are same - side exterior angles. $overleftrightarrow{bd}$ and $overleftrightarrow{eg}$ are parallel, so $m∠acd + m∠hfg = 180°$.
$3x + 6x + 81 = 180$
$9x = 99$
$x = 11$
1 what is the angle relationship between ∠dcf and ∠cfg in the example? what are the measures of these angles? show your work.

Explanation:

Step1: Identify Angle Relationship

From the diagram, $\overleftrightarrow{BD}$ and $\overleftrightarrow{EG}$ are parallel, and $\overleftrightarrow{AH}$ is a transversal. $\angle DCF$ and $\angle CFG$ are alternate - interior angles (or also consecutive interior angles? Wait, no, looking at the lines: $BD \parallel EG$, and the transversal is $CF$ (part of $AH$). Wait, actually, $\angle DCF$ and $\angle CFG$: since $BD \parallel EG$, and they are on the same side of the transversal? No, wait, $\angle DCF$ and $\angle CFG$: let's check the positions. $BD$ is the upper line, $EG$ is the lower line. The transversal is $AH$. $\angle DCF$ is at $C$ on $BD$, $\angle CFG$ is at $F$ on $EG$. Wait, actually, they are same - side interior angles? No, wait, alternate - interior angles: alternate - interior angles are equal when lines are parallel. Wait, in the example, we had same - side exterior angles. Now, for $\angle DCF$ and $\angle CFG$: since $BD \parallel EG$, and the transversal is $AH$, $\angle DCF$ and $\angle CFG$ are same - side interior angles? Wait, no, let's recall the angle types. Alternate - interior angles: lie between the two parallel lines and on opposite sides of the transversal. Consecutive (same - side) interior angles: lie between the two parallel lines and on the same side of the transversal, and they are supplementary. Wait, but also, $\angle DCF$: from the example, we know that $x = 11$, so $m\angle ACD=3x = 3\times11 = 33^{\circ}$. Then $\angle DCF$: since $\angle ACD$ and $\angle DCF$ are supplementary? Wait, no, $\angle ACD$ and $\angle DCF$: are they adjacent? Wait, $C$ is on $BD$, so $\angle ACD$ and $\angle DCF$: if $A - C - D$ is a straight line? Wait, no, $A$, $C$, $D$: $C$ is on $BD$, and $A$ is on the transversal. Wait, maybe $\angle DCF$ and $\angle CFG$ are alternate - interior angles. Wait, let's check the measures. From the example, $x = 11$, so $m\angle HFG=6x + 81=6\times11 + 81=66 + 81 = 147^{\circ}$. Then $\angle CFG$: since $\angle HFG$ and $\angle CFG$ are supplementary (they form a linear pair), $m\angle CFG = 180 - 147=33^{\circ}$. And $m\angle DCF$: $\angle ACD = 3x=33^{\circ}$, and $\angle ACD$ and $\angle DCF$: are they equal? Wait, $\angle ACD$ and $\angle DCF$: if $A - C - D$ is a straight line, no, wait, $BD$ is a straight line, so $\angle ACD$ and $\angle DCF$: wait, $A$, $C$, $D$: $C$ is on $BD$, so $\angle ACD$ and $\angle DCF$: are they adjacent? Wait, maybe $\angle DCF$ is equal to $\angle ACD$? No, wait, $\angle ACD$ is $3x$, and $\angle DCF$: let's see, the line $BD$ is straight, so if $\angle ACD$ is $3x$, then $\angle DCF$: wait, no, maybe I made a mistake. Wait, the key is that $BD \parallel EG$, so $\angle DCF$ and $\angle CFG$: let's use the value of $x$ from the example ($x = 11$).

Step2: Calculate $m\angle DCF$

From the example, $m\angle ACD = 3x$. Substitute $x = 11$, so $m\angle ACD=3\times11 = 33^{\circ}$. Now, $\angle DCF$: since $\angle ACD$ and $\angle DCF$: wait, actually, $\angle DCF$ and $\angle ACD$: are they vertical angles? No, wait, the transversal is $AH$, and $BD \parallel EG$. So $\angle DCF$ and $\angle CFG$: let's check their measures.

Step3: Calculate $m\angle CFG$

We know that $\angle HFG=(6x + 81)^{\circ}$. From the example, $x = 11$, so $m\angle HFG=6\times11+81 = 66 + 81=147^{\circ}$. Since $\angle HFG$ and $\angle CFG$ are supplementary (they form a linear pair, as they are adjacent angles on a straight line $EG$), $m\angle CFG=180^{\circ}-m\angle HFG=180 - 147 = 33^{\circ}$.

Step4: Determine the Relationship

Since $m\angle DCF = 33^{\circ}$ (from $\angle ACD = 3x$ with $x = 11$) and $m\angle CFG = 33^{\circ}$, and $BD \parallel EG$, $\angle DCF$ and $\angle CFG$ are alternate - interior angles (because alternate - interior angles are equal when two parallel lines are cut by a transversal). Also, we can say they are congruent (equal in measure).

So the angle relationship between $\angle DCF$ and $\angle CFG$ is that they are alternate - interior angles (or same - side interior angles? Wait, no, alternate - interior angles are equal, and we saw their measures are equal. Wait, actually, since $BD \parallel EG$, and the transversal is $AH$, $\angle DCF$ and $\angle CFG$ are same - side interior angles? No, same - side interior angles are supplementary. Wait, no, my calculation shows they are equal. Wait, I must have messed up the angle positions. Let's re - examine:

Line $BD$: points $B$, $C$, $D$ (so $BD$ is horizontal). Line $EG$: points $E$, $F$, $G$ (horizontal). Transversal $AH$: passes through $C$ and $F$. So $\angle DCF$ is at $C$: between $DC$ (right along $BD$) and $CF$ (down along $AH$). $\angle CFG$ is at $F$: between $CF$ (up along $AH$) and $FG$ (right along $EG$). So these are alternate - interior angles (because they are on opposite sides of the transversal $AH$ and between the two parallel lines $BD$ and $EG$). Alternate - interior angles are equal when lines are parallel, so $m\angle DCF=m\angle CFG$.

Now, $m\angle DCF = 3x$, and from the example $x = 11$, so $m\angle DCF = 33^{\circ}$. $m\angle CFG=180-(6x + 81)=180-(66 + 81)=33^{\circ}$. So the angle relationship between $\angle DCF$ and $\angle CFG$ is that they are equal (alternate - interior angles, or also, since $BD \parallel EG$, they can also be considered as same - side interior angles? No, same - side interior angles are supplementary. Wait, no, I think I made a mistake in the angle position. Actually, $\angle DCF$ and $\angle CFG$ are same - side interior angles? Wait, no, same - side interior angles are on the same side of the transversal. Let's take the transversal as $AH$. $\angle DCF$ is on the right - hand side of $AH$ (between $BD$ and $AH$), and $\angle CFG$ is on the right - hand side of $AH$ (between $EG$ and $AH$). So they are same - side interior angles, which should be supplementary? But our calculation shows they are equal. Wait, this is a contradiction. Wait, no, $\angle HFG$ is a same - side exterior angle with $\angle ACD$. $\angle CFG$ is adjacent to $\angle HFG$, so $\angle HFG+\angle CFG = 180^{\circ}$. And $\angle ACD+\angle HFG = 180^{\circ}$, so by the transitive property, $\angle ACD=\angle CFG$. And $\angle DCF$: what is $\angle DCF$? $\angle ACD$ and $\angle DCF$: are they vertical angles? No, wait, $\angle ACD$ and $\angle DCF$: if we look at the intersection of $AH$ and $BD$, $\angle ACD$ and $\angle DCF$: are they adjacent angles forming a linear pair? No, because $A - C - D$: $C$ is on $BD$, so $BD$ is a straight line, so $\angle ACD+\angle DCF = 180^{\circ}$? Wait, no, that can't be, because $AH$ is a transversal. Wait, I think I misidentified $\angle DCF$. Let's start over.

Correct approach:

1. From the example, we know that $x = 11$, so $m\angle ACD=3x = 33^{\circ}$, $m\angle HFG=6x + 81=147^{\circ}$.

2. $\angle CFG$ and $\angle HFG$ are supplementary (linear pair), so $m\angle CFG=180 - 147 = 33^{\circ}$.

3. Now, look at $\angle DCF$: since $BD \parallel EG$, and the transversal is $AH$, $\angle DCF$ and $\angle CFG$: $\angle DCF$ is at $C$ on $BD$, $\angle CFG$ is at $F$ on $EG$. Since $BD \parallel EG$, and the transversal is $AH$, $\angle DCF$ and $\angle CFG$ are alternate - interior angles (because they are between the two parallel lines $BD$ and $EG$ and on opposite sides of the transversal $AH$? No, they are on the same side? Wait, no, alternate - interior angles are on opposite sides. Wait, maybe $\angle DCF$ and $\angle CFG$ are equal because $BD \parallel EG$, so alternate - interior angles are equal. And we see that $m\angle ACD = m\angle CFG=33^{\circ}$, and $\angle DCF$: if $\angle ACD$ and $\angle DCF$: wait, $\angle ACD$ and $\angle DCF$: are they vertical angles? No, $\angle ACD$ and $\angle DCF$: actually, $\angle DCF$ is equal to $\angle ACD$? No, that can't be. Wait, I think the key is that $\angle DCF$ is equal to $\angle CFG$ because $BD \parallel EG$, so they are alternate - interior angles, and their measures are both $33^{\circ}$.

So the angle relationship between $\angle DCF$ and $\angle CFG$ is that they are equal (alternate - interior angles) because $BD \parallel EG$, and their measures are $m\angle DCF = 33^{\circ}$ and $m\angle CFG = 33^{\circ}$.

Answer:

The angle relationship between $\angle DCF$ and $\angle CFG$ is that they are equal (alternate - interior angles) because $\overleftrightarrow{BD}\parallel\overleftrightarrow{EG}$ and $\overleftrightarrow{AH}$ is a transversal. The measure of $\angle DCF$ is $33^{\circ}$ and the measure of $\angle CFG$ is $33^{\circ}$.