Verifying Properties of Triangles and Quadrilaterals

Triangles are classified by their sides: scalene (no equal sides), isosceles (at least two equal sides) and equilateral (all three sides equal). They are also classified by their angles, and a right triangle has one $90^{\circ }$ angle. To classify a triangle from its vertices, find all three side lengths; to test for a right angle, find the slopes of the sides and look for a pair with product $-1$ (or one horizontal side and one vertical side). Another test for a right angle uses the side lengths: if the squares of the two shorter sides add up to the square of the longest side, the triangle is right-angled (this is the converse of the Pythagorean theorem). For $A(-2,3)$, $B(2,5)$ and $C(4,1)$: $A{B}^2=16+4=20$, $B{C}^2=4+16=20$ and $A{C}^2=36+4=40$. Two sides are equal, so the triangle is isosceles; and $20+20=40$, so it is also right-angled, with the right angle at $B$, opposite the longest side.

The diagonals (segments joining opposite vertices) carry useful properties too. In every parallelogram the diagonals bisect each other: they share the same midpoint. In a rectangle the diagonals also have equal length, and in a rhombus they are also perpendicular; a square has all three properties. These facts also work backwards: if a quadrilateral's diagonals bisect each other, it is a parallelogram; if they bisect each other and are equal, it is a rectangle; if they bisect each other and are perpendicular, it is a rhombus. So you can tell whether a quadrilateral is a parallelogram, rectangle, rhombus or square by looking only at its diagonals.

Classifying a quadrilateral on the coordinate plane | Analytic geometry | Geometry | Khan Academy

A midsegment of a triangle joins the midpoints of two of its sides. The midsegment theorem says that a midsegment is parallel to the third side and half as long. Test it on $A(-3,-2)$, $B(5,2)$ and $C(1,6)$. The midpoint of $AC$ is $M(-1,2)$ and the midpoint of $BC$ is $N(3,4)$. The slope of $MN$ is $\frac{4-2}{3-(-1)}=\frac{2}{4}=\frac{1}{2}$, and the slope of $AB$ is $\frac{2-(-2)}{5-(-3)}=\frac{4}{8}=\frac{1}{2}$, so $MN\parallel AB$ (the symbol $\parallel$ means "is parallel to"). The lengths are $MN=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$ and $AB=\sqrt{64+16}=\sqrt{80}=4\sqrt{5}$, so $AB=2\times MN$. Exact radicals make the comparison clean: the decimals $4.5$ and $8.9$ are rounded and only roughly show the factor of $2$.

The Triangle Midsegment Theorem

A median of a triangle joins a vertex to the midpoint of the opposite side, so every triangle has three medians. In the triangle above, the median from $A(-3,-2)$ goes to $N(3,4)$, the midpoint of $BC$; its slope is $\frac{4-(-2)}{3-(-3)}=\frac{6}{6}=1$, and substituting $A$ into $y=x+b$ gives $b=1$, so its equation is $y=x+1$. Investigating many triangles shows that the three medians always pass through one point, the centroid, the balance point of a triangle cut from cardboard. Its coordinates are the averages of the three vertices: here $\left(\frac{-3+5+1}{3},\frac{-2+2+6}{3}\right)=(1,2)$, and indeed $1+1=2$, so it lies on $y=x+1$. Do not confuse a median with an altitude (a segment from a vertex perpendicular to the opposite side) or with a perpendicular bisector; they coincide only in special cases, such as the median to the base of an isosceles triangle.

1. State the goal in terms you can calculate: "rectangle" means "opposite sides parallel and one right angle"; "diagonals bisect each other" means "same midpoint".
2. Sketch the figure on a grid and label the vertices in order around the shape. The sketch guides you; it does not prove anything.
3. Calculate each slope, length or midpoint you need, showing the substitution.
4. Compare the results, using exact values (fractions and radicals) rather than rounded decimals.
5. Conclude with a sentence that names the property and the evidence, such as "Since both diagonals have midpoint $(0.5,1.5)$, they bisect each other."