Geometry Lesson 7C: Parallel Line Theorems PDF

Here is the markdown conversion of the text from the images you sent: # GEOMETRY - LESSON 7C ## MORE PARALLEL LINE THEOREMS ### THEOREM If two lines are parallel, then all points on one line are equidistant from the other line. **GIVEN**: $m || n; AC \perp n; BD \perp n$ A and B are any points on...

Here is the markdown conversion of the text from the images you sent: # GEOMETRY - LESSON 7C ## MORE PARALLEL LINE THEOREMS ### THEOREM If two lines are parallel, then all points on one line are equidistant from the other line. **GIVEN**: $m || n; AC \perp n; BD \perp n$ A and B are any points on m **PROVE**: $AC=BD$ **PROOF**: Since $AB$ and $CD$ are contained in parallel lines, $AB || CD$. We also know that $AC || BD$ since they are both coplanar and perpendicular to line $n$. Therefore, the quadrilateral $ABDC$ is a parallelogram. Since opposite sides of a parallelogram are congruent, $AC = BD$. # MORE THEOREMS INVOLVING PARALLEL LINES ### THEOREM If three parallel lines cut congruent segments on one transversal, then they cut congruent segments on every transversal. **GIVEN**: $AX || BY || CZ$ $AB = BC$ **PROVE**: $XY=YZ$ **PROOF**: Draw $XR$ and $YS$ parallel to $AC$ as shown, creating parallelograms $AXRB$ and $BYSC$. Since opposite sides of a parallelogram are congruent, $XR \cong AB$ and $BC \cong YS$. Using the given $AB = BC$, we know that $XR \cong YS$ by the transitive property. Parallel lines are cut by transversals to so that $\angle 1 \cong \angle 3$, $\angle 3 \cong \angle 4$, and $\angle 4 \cong \angle 2$. Therefore $\angle 1 \cong \angle 2$ by the transitive property. Also, $\angle 5 \cong \angle 6$ because they are alternate interior $\angle s$ on parallel lines. Therefore, $\triangle XYR \cong \triangle YZS$ by $AAS$ and $XY \cong YZ$ by $CPCTC$. # MORE THEOREMS INVOLVING PARALLEL LINES ### THEOREM A line that contains the midpoint of one side of a triangle and is parallel to another side passes through the midpoint of the third side. **GIVEN**: $M$ is the midpoint of $AB$. $MN || BC$ **PROVE**:$N$ is the midpoint of $AC$ **PROOF**: Draw $AD$ parallel to $MN$. Then $AD, MN$, and $BC$ are three parallel lines that cut congruent segments on transversal $AB$. Therefore, the lines will also cut congruent segments on $AC$. So $AN \cong NC$ and $N$ is the midpoint of $AC$. # MORE THEOREMS INVOLVING PARALLEL LINES ## MIDSEGMENT: a segment that joins two midpoints of a polygon ### THEOREM A midsegment of a triangle is parallel to the third side and half as long. **GIVEN**: $MN$ is a midsegment of $\triangle{ABC}$ **PROVE**: (1) $MN || BC$ (2) $MN = \frac{1}{2}BC$ **PROOF (PART 1)**: There exists only one line going through $M$ parallel to $BC$. Since that line must pass through $N$, the midpoint of the third side, $MN || BC$. ## MORE THEOREMS INVOLVING PARALLEL LINES ### THEOREM A midsegment of a triangle is parallel to the third side and half as long. **GIVEN**: $MN$ is a midsegment of $\triangle{ABC}$ **PROVE**: (1) $MN || BC$ (2) $MN = \frac{1}{2}BC$ **PROOF (PART 2)**: Let $L$ be the midpoint of $BC$ and draw $NL$. Since $NL$ crosses through the midpoint of two sides of the triangle it is parallel to the third side, $MB$. So $NL || AB$ and since we've already proven that $MN || BC$, MNLB is a parallelogram. Since its opposite sides are congruent, $MN = BL$. Since $L$ is the midpoint of $BC$, $BL = \dfrac{1}{2}BC$ and therefore $MN = \dfrac{1}{2}BC$ by the transitive property.