Download Right Triangles: Properties and Relationships and more Lecture notes Geometry in PDF only on Docsity!
Lecture 21: Right Triangles
21.1 Right Triangles
Definition We call a triangle 4 ABC in a protractor geometry a right triangle if one of ∠A, ∠B, or ∠C is a right angle. We call a side opposite a right angle in a right triangle a hypotenuse.
Definition In a protractor geometry, we call AB the longest side of 4 ABC if AB > AC and AB > BC, and we call AB a longest side if AB ≥ AC and AB ≥ BC.
Theorem If, in a neutral geometry, 4 ABC is a right triangle with ∠C a right angle, then ∠A and ∠B are acute and AB is the longest side of 4 ABC.
Proof Let D be a point such that D − C − B. Then ∠DCA is a right angle, exterior to 4 ABC with remote interior angles ∠A and ∠B. Hence ∠DCA > ∠A and ∠DCA > ∠B. Since ∠DCA ' ∠ACB, it follows that ∠C > ∠A and ∠C > ∠B. Hence ∠A and ∠B are both acute. Moreover, it also follows that AB > AC and AB > BC.
Note that the previous result says that a right triangle has only one hypotenuse and that the hypotenuse is the longest side of a right triangle.
Definition Given a right triangle 4 ABC with right angle at C in a neutral geometry, we call AC and BC the legs of 4 ABC.
Theorem If P and Q are points and is a line in a neutral geometry with Q ∈ and
P /∈ L, then
←→ P Q ⊥ if and only if P Q ≤ P R for all R ∈.
Proof Suppose
←→ P Q ⊥ and R ∈. If R = Q, then P Q = P R. If R 6 = Q, then 4 P QR is a right triangle with hypotenuse P R. Hence P Q < P R. Thus, in either case, P Q ≤ P R.
Now suppose P Q ≤ P R for all R ∈ . Let S be the point on such that P S is the unique line through P which is perpendicular to `. If S 6 = Q, then 4 P QS is a right triangle with hypotenuse P Q, from which it follows that P S < P Q, contradicting our assumption about
Q. Hence we must have S = Q, and so
←→ P Q ⊥ `.
Definition Given a line ` and a point P in a neutral geometry {P, L, d, m}, we call
d(P, `) =
d(P, Q), if P /∈ and Q is the unique point on for which
←→ P Q ⊥ , 0 , if P ∈,
the distance from P to `.
Lecture 21: Right Triangles 21-
Note that for any P ∈ , d(P,) ≤ d(P, Q) for all Q ∈ , with d(P,) = d(P, Q) if and only
if
←→ P Q ⊥ `.
Definition Given 4 ABC in a neutral geometry with D being the unique point on
←→ AB
for which
←→ CD ⊥
←→ AB, we call CD the altitude from C and we call D the foot of the altitude from C.
Theorem If, in a neutral geometry, AB is the longest side of 4 ABC and D is the foot of the altitude from C, then A − D − B.
Proof Suppose D − A − B. Then CB is the hypotenuse of the right triangle 4 CBD, and so CB > BD. But DB > AB, so CB > AB, contradicting the assumption that AB is the longest side of 4 ABC.
Now suppose D = A. Then 4 ABC is right triangle with right angle at A, and so BC > AB, again contradicting the assumption that AB is the longest side of 4 ABC.
Similarly, we cannot have A − B − D nor D = B. Hence A − D − B.
Hypotenuse-Leg (HL) If, in a neutral geometry, 4 ABC and 4 DEF are right triangles with right angles at C and F , AB ' DE, and AC ' DF , then 4 ABC ' 4DEF.
Note that we could prove Hypotenuse-Leg in the Euclidean Plane using the Pythagorean Theorem and Side-Side-Side.
Proof Let G be the point on
←→ EF such that E − F − G and F G ' BC. Then AC ' DF , ∠ACB ' ∠DF G (they are both right angles), and CB ' F G, and so 4 ACB ' 4DF G by Side-Angle-Side. In particular, AB ' DG, so DG ' DE. Hence 4 DEG is isosceles, and so ∠DEF ' ∠DGF. Thus 4 DEF ' 4DGF by Side-Angle-Angle. Hence 4 DEF ' 4 ABC.
Hypotenuse-Angle (HA) If, in a neutral geometry, 4 ABC and 4 DEF are right triangles with right angles at C and F , AB ' DE, and ∠A ' ∠D, then 4 ABC ' 4DEF.
Proof See homework.
Theorem If, in a neutral geometry {P, L, d, m}, is the perpendicular bisector of AB, then = {P : P ∈ P, AP = BP }.
Proof Let P be a point with AP = BP. If P ∈
←→ AB, then P is the midpoint of AB, and
so P ∈ `. So suppose P /∈
←→ AB. Let N be the unique point on
←→ AB such that
←→ P N ⊥
←→ AB.
