Geometry Stations Review: Triangle Properties, Midsegments, and Inequalities

Practice vocabulary and concept flashcards covering angle bisectors, perpendicular bisectors, triangle midsegments, angle ordering, hinge theorem comparison, and triangle inequality rules from Stations 2 through 6.

Angle Bisector

A ray or segment that divides an angle into two congruent angles, such as segment $AD$ bisecting $\rule{0pt}{0pt}\text{angle } BAC$ so that $m\angle BAD = m\angle CAD$.

Perpendicular Bisector

A segment or line that intersects a side of a triangle at its midpoint at a $90^\circ$ angle, dividing that side into two equal halves.

Triangle Midsegment Theorem

A theorem stating that a segment connecting the midpoints of two sides of a triangle is parallel to the third side and half as long as that third side, expressed as $MN = \frac{1}{2}SR$ or $SR = 2 \times MN$.

Triangle Inequality Theorem

A theorem stating that three side lengths can form a triangle if and only if the sum of any two side lengths is strictly greater than the third side length ($a + b > c$).

Hinge Theorem

A theorem used to compare angle measures and opposite side lengths between two triangles that share two congruent corresponding sides.

Station 2: Angle Bisector Calculation for $m\angle ADC$

In $\triangle ABC$, given $m\angle B = 47^\circ$, $m\angle C = 31^\circ$, and $AD$ bisects $\angle BAC$: $m\angle BAC = 180^\circ - (47^\circ + 31^\circ) = 102^\circ$, making $m\angle CAD = 51^\circ$ and $m\angle ADC = 180^\circ - (51^\circ + 31^\circ) = 98^\circ$.

Station 2: Perpendicular Bisector Calculation for $m\angle CAD$

Given $m\angle C = 33^\circ$ and $AD$ is the perpendicular bisector of $BC$: $AD \perp BC$ implies $m\angle ADC = 90^\circ$, so $m\angle CAD = 180^\circ - 90^\circ - 33^\circ = 57^\circ$.

Station 2: Perpendicular Bisector Calculation for $DC$

Given $BC = 22$ and $AD$ is the perpendicular bisector of $BC$: $AD$ intersects $BC$ at its midpoint $D$, yielding $DC = \frac{BC}{2} = \frac{22}{2} = 11$.

Station 3: Angle Ordering Method

The procedure of listing interior angles from least to greatest by using coordinate lengths, where the smallest angle is always opposite the shortest side and the largest angle is opposite the longest side.

Station 4: Midsegment Length $SR$ with Algebraic Expressions

Given midsegment $MN = 2x$ and side $SR = 3x + 7$: using $SR = 2(MN)$ gives $3x + 7 = 2(2x) \implies 3x + 7 = 4x \implies x = 7$, which yields $SR = 3(7) + 7 = 28$.

Station 4: Midsegment Length $AF$ with Algebraic Expressions

Given base $AC = 3x - 9$ and midsegment $ED = x + 2$: solving $3x - 9 = 2(x + 2)$ gives $x = 13$, so $AC = 30$ and $AF = \frac{AC}{2} = 15$.

Station 5: Triangle Feasibility for Side Lengths $(2, 9, 12)$

Side lengths $2$, $9$, and $12$ cannot form a triangle because $2 + 9 = 11$, which is not greater than $12$ ($11 \ngtr 12$).

Station 5: Triangle Feasibility for Side Lengths $(18, 32, 14)$

Side lengths $18$, $32$, and $14$ cannot form a triangle because $14 + 18 = 32$, which equals the third side rather than being strictly greater ($32 \ngtr 32$).

Station 5: Triangle Feasibility for Side Lengths $(103, 41.9, 62.5)$

Side lengths $103$, $41.9$, and $62.5$ can form a triangle because the sum of the two shorter sides $41.9 + 62.5 = 104.4$ is strictly greater than $103$ ($104.4 > 103$).

Station 5: Triangle Feasibility for Side Lengths $(28, 28, 28)$

Side lengths $28$, $28$, and $28$ can form an equilateral triangle because $28 + 28 = 56$, which is strictly greater than $28$ ($56 > 28$).