4.5 Trapezoids
We now turn our attention to trapezoids, and to a segment in a trapezoid that is analogous to the midline of a triangle.
- Definition: A trapezoid is a quadrilateral with exactly one pair of opposite sides parallel.
- Theorem 4.15: The base angles of an isosceles trapezoid are congruent.
Given: $ABCD$ is an
isosceles trapezoid with $\overline{AB}\parallel\overline{CD}$. Prove: $\angle A\cong \angle B$ and $\angle C \cong \angle D$ |
Proof: (outline): By dropping perpendiculars from $A$ and $B$ to the base $\overline{CD}$, congruent triangles (using Hypotenuse-Leg) can be formed. That leads to the base angles being congruent.
- Definition: The median of a trapezoid is the segment joining the midpoints of the legs of the trapezoid.
- Theorem 4.16: The median of a trapezoid is parallel to the bases of the trapezoid and the length of the median is equal to half the sum of the lengths of the bases.
Given: $ABCD$ is a
trapezoid with median $\overline{EF}$. Prove: a. $\overline{EF}\parallel\overline{BC}$, and $\overline{EF}\parallel\overline{AD}$; b. $EF=\frac{1}{2}(BD+AD)$ |
Proof: (Part a) Introduce a line through $F$ that is parallel to $\overline{AB}$. Since $F$ is the midpoint of $\overline{CD}$, we have $\overline{CF}\cong\overline{FD}$. $\angle CFP \cong \angle DFR$ and $\angle PCF \cong \angle RDF$. Therefore $\triangle PCF \cong \triangle RDF$ by ASA. Since the opposite sides of $ABPR$ are parallel, $ABPR$ is a parallelogram; hence $AB=RP$. $E$ is the midpoint of $\overline{AB}$, so $AE=\frac{1}{2}AB$. Since $\triangle PCF \cong \triangle RDF$, we have $\overline{RF}\cong\overline{PF}$ and $RF=\frac{1}{2}RP$. By the Transitive Property, $AE=RF$. Since $\overline{AE}\parallel\overline{RF}$, quadrilateral $AEFR$ is a parallelogram. This makes $\overline{EF}\parallel\overline{ARD}$ and therefore also parallel to $\overline{BCP}$ since a segment parallel to one of two parallel lines is parallel to the other.
(Part b) Now introduce diagonal $\overline{BD}$. Since $\overline{EQ}$ passes through the midpoint of one side of $\triangle ADB$ and is parallel to a second side, it must pass through the midpoint of the third side (Theorem 4.15). This means that $\overline{EQ}$ is a midline of $\triangle ADB$ and thus $EQ=\frac{1}{2}AD$. Since $Q$ and $F$ are midpoints of $\overline{BD}$ and $\overline{CD}$, respectively, $QF=\frac{1}{2}BC$. Adding these two results, we have $EQ+QF=\frac{1}{2}AD+\frac{1}{2}BC=\frac{1}{2}(AD+BC)$.