Proving Lines Parallel or Perpendicular: Practice

If the lines are parallel, the forward corresponding-angle theorem forces the measures to be equal. Thus every suitable value must satisfy

$5x-11=3x+27$

Subtract $3x$ and add 11 to get $2x=38$, so $x=19$ is the only solution. At this value both measures are $84^\circ$, which is valid. Conversely, their equality establishes parallelism by the corresponding-angle converse.

If the lines are parallel, the forward same-side interior theorem forces a supplementary pair. Every suitable value must therefore satisfy

$(2x+17)+(4x+13)=180$

Combining terms gives $6x+30=180$, then $6x=150$ and the unique solution $x=25$. The measures become $67^\circ$ and $113^\circ$. Conversely, these valid supplementary angles establish parallelism by the same-side interior converse.

If the lines are parallel, the forward alternate exterior theorem forces equal measures. Every suitable value must satisfy

$3x+4=5x-30$

Adding 30 and subtracting $3x$ gives $34=2x$, so the unique solution is $x=17$. Substitution gives $55^\circ$ for each. Conversely, these valid congruent alternate exterior angles establish parallelism by the converse.

The perpendicular transversal theorem makes $t$ perpendicular to $m$. The angle must be right.

$8x-22=90$

Add 22 and divide by 8 to obtain $x=14$. Substitution gives $90^\circ$.

At $x=50$, each expression gives a measure outside the valid range.

$50+200=250$

Each numbered sector at a crossing of distinct lines measures strictly between $0^\circ$ and $180^\circ$. The algebraic equality therefore fails the geometric conditions.

Line $n$ meets $m$ at a right angle. Since $\ell\parallel m$ and all three lines are coplanar, the proved crossing result shows that $n$ also meets $\ell$, making $n$ a transversal.

The perpendicular transversal theorem now gives $n\perp\ell$.

Angle 5 corresponds to angle 1 at the other crossing. Its stated $102^\circ$ measure would give a congruent corresponding pair.

The other facts follow locally at the upper crossing and do not constrain $m$.

An angle measure and a starting ray can give rays on opposite sides.

Choosing the specified side places the copied angle in the required corresponding position. Equality of measures without the angle positions would not justify the intended converse.

The angle-measure foundation fixes a unique ray on a chosen side for an opening strictly between $0^\circ$ and $180^\circ$.

The selected rays of $n$ and $\ell$ coincide, so their full lines are the same. Because $n$ is parallel to the original line, so is $\ell$.

The perpendicular segment supplies the point-to-line distance, so $PH=32$.

$5k-3=32$

Add 3 and divide by 5 to get $k=7$. Then $5(7)-3=32$.

If $\ell$ and $m$ intersected at a point outside $n$, two distinct parallels to $n$ would pass through that point.

That contradicts the accepted uniqueness of a parallel. The distinct coplanar lines therefore do not meet and are parallel.

A protractor has limited precision, and the readings were explicitly approximate.

The converse requires congruent angles, an exact condition. A measurement can check or suggest the drawing without proving that condition.