Proving Lines Parallel or Perpendicular: Practice

If the lines are parallel, the forward same-side exterior theorem forces the angle sum to be $180^\circ$. Every suitable value must satisfy

$(7x-5)+(2x+14)=180$

This gives $9x+9=180$, then $9x=171$ and the unique solution $x=19$. The measures are $128^\circ$ and $52^\circ$, both valid. Conversely, their sum establishes parallelism by the same-side exterior converse.

Vertical angles 1 and 3 are congruent, so equate their measures.

$3x+11=7x-49$

Subtracting $3x$ and adding 49 gives $60=4x$, hence $x=15$. Then $m\angle1=56^\circ$, equal to the given $m\angle5$. Angles 1 and 5 are corresponding, so their converse gives $\ell\parallel m$.

Angles 3 and 6 are same-side interior angles.

$121^\circ+59^\circ=180^\circ.$

The same-side interior converse proves parallelism; congruence is not required for this test.

At $R$, choose one side of $n$ and one ray of $n$ as the starting ray. Only one ray on that side makes a $90^\circ$ opening.

Both claimed perpendiculars use that ray and its opposite, so they are the same full line. This contradicts their claimed distinctness.

Since each proposed line is parallel to $c$, neither meets $c$; their assumed common point $R$ is outside $c$.

Both would be different parallels to $c$ through $R$, contradicting the accepted uniqueness of a parallel.

Congruence by itself does not identify the geometric relationship between the openings.

For example, congruent same-side interior angles of $38^\circ$ would sum to $76^\circ$, which fails the required supplementary condition. A valid test needs the angle positions.

The order $A,P,C$ makes $PC$ the continuation of the same transversal direction as $AP$. The specified side puts the new ray in the corresponding position relative to $AB$.

Thus the copied angles are a congruent corresponding pair at distinct crossings, giving parallelism by the converse.

Equate the congruent corresponding measures.

$4z+8=6z-36$

This gives $44=2z$, so $z=22$. Both angle measures are $96^\circ$. Their congruence proves $\ell\parallel m$, while $96^\circ$ is not right, so $t$ is not perpendicular.

Angles 1 and 2 form a linear pair, so their measures sum to $180^\circ$. They are also congruent, so each measure is

$180^\circ\div2=90^\circ$

Thus $t\perp\ell$. The other given says angle 5 is congruent to angle 1, so angles 1 and 5 are a congruent corresponding pair. Their converse proves $\ell\parallel m$, and the right angle at the lower crossing also gives $t\perp m$.

Because $H$ is the perpendicular foot, the distance definition gives $PH=46$ cm. Since $X$ lies between $P$ and $H$, segment addition gives

$(4w-3)+(2w+7)=46$

Combine terms to get $6w+4=46$, so $6w=42$ and $w=7$. Now substitute into the requested part length.

$PX=4(7)-3=25$

The other part is $XH=2(7)+7=21$ cm, and $25+21=46$ cm checks the whole perpendicular length.

Angles 4 and 5 are same-side interior angles. Their measures satisfy

$47^\circ+133^\circ=180^\circ.$

Their converse proves parallelism. Each other pair describes vertical angles or a linear pair at one crossing and leaves the other line direction unrestricted.

Since $a\parallel b$ and $\ell$ crosses $a$ perpendicularly, the proved crossing result shows that $\ell$ also crosses $b$. It is therefore a transversal of both parallels, and the perpendicular transversal theorem gives $\ell\perp b$.

Both distinct coplanar lines $\ell$ and $m$ are now perpendicular to $b$, so the two-perpendiculars result gives $\ell\parallel m$.