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From Wikipedia

Right triangle

A right triangle (American English) or right-angled triangle (British English) is a triangle in which one angle is a right angle (that is, a 90 degree angle). The relation between the sides and angles of a right triangle is the basis for trigonometry.

Terminology

The side opposite the right angle is called the hypotenuse(side c in the figure above). The sides adjacent to the right angle are called legs (or catheti, singular:cathetus). Side a may be identified as the side adjacent to angle B and opposed to (or opposite) angle A, while side b is the side adjacent to angle A and opposed to angle B.

Principal properties

Area

As with any triangle, the area is equal to one half the base multiplied by the corresponding height. In a right triangle, if one leg is taken as the base then the other is height, so the area of a right triangle is one half the product of the two legs. As a formula:

\text{Area}=\tfrac{1}{2}ab

where a and b are the legs of the triangle.

If the incircle is tangent to the hypotenuse AB at point P, then PA = s– a and PB = s– b, and the area is given by

\text{Area}=\text{PA} \cdot \text{PB} = (s-a)(s-b).

Altitude

If an altitude is drawn from the vertex with the right angle to the hypotenuse then the triangle is divided into two smaller triangle which are both similar to the original and therefore similar to each other. From this:

  • The altitude is the mean proportional of the two segments of the hypotenuse.
  • Each leg of the triangle is the mean proportional of the hypotenuse and the adjacent segment.

In equations,

f^2=de,\,b^2=ce,\,a^2=cd

where a, b, c, d, e, f are as shown in the diagram. Thus fc = ab.

Moreover, the altitude to the hypotenuse is related to the legs of the right triangle by

\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{ f^2}.

Pythagorean theorem

The Pythagorean theorem states that:

In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).

This can be stated in equation form as c2 = a2 + b2 where c is the length of the hypotenuse and a and b are the lengths of the remaining two sides.

Trigonometric ratios

The trigonometric functions for acute angles can be defined as ratios of the sides of a right triangle. For a given angle, a right triangle may be constructed with this angle, and the sides labeled opposite, adjacent and hypotenuse with reference to this angle according to the definitions above. These ratios of the sides do not depend on the particular right triangle chosen, but only on the given angle, since all triangles constructed this way are similar. If, for a given angle α, the opposite side, adjacent side and hypotenuse are labeled O, A and H respectively, then the trigonometric functions are

\sin\alpha =\frac {O}{H},\,\cos\alpha =\frac {A}{H},\,\tan\alpha =\frac {O}{A},\,\sec\alpha =\frac {H}{A},\,\cot\alpha =\frac {A}{O},\,\csc\alpha =\frac {H}{O}.

Special right triangles

The values of the trigonometric functions can be evaluated exactly for certain angles using right triangle with special angles. These include the 30-60-90 triangle which can be used to evaluate the trigonometric functions for any multiple of π/6, and the 45-45-90 triangle which can be used to evaluate the trigonometric functions for any multiple of π/4.

Thales' theorem

Thales' theorem states that if A is any point of the circle with diameter BC (except B or C themselves) â–³ABC is a right triangle with A the right angle. The converse states that if a right triangle is inscribed in a circle then the hypotenuse will be a diameter of the circle. A corollary is that the length of the hypotenuse is twice the distance from the right angle vertex to the midpoint of the hypotenuse. Also, the center of the circle that circumscribes a right triangle is the midpoint of the hypotenuse and its radius is one half the length of the hypotenuse.

Medians

The following formulas hold for the medians of a right triangle:

m_a^2 + m_b^2 = 5m_c^2 = \frac{5}{4}c^2.

The median on the hypotenuse of a right triangle divides the triangle into two isosceles triangles, because the median equals one-half the hypotenuse.

Relation to various means and the golden ratio

Let H, G, and A be the harmonic mean, the geometric mean, and the arithmetic mean of two positive numbers a and b with a> b. If a right triangle has legs H and G and hypotenuse A, then

\frac{A}{H} = \frac{A^{2}}{G^{2}} = \frac{G^{2}}{H^{2}} = \phi \,

and

\frac{a}{b} = \phi^{3}, \,

where \phi is the golden ratio \tfrac{1+ \sqrt{5}}{2}. \,

Other properties

The radius of the incircle of a right triangle with legs a and b and hypotenuse c is

r = \frac{1}{2}(a+b-c) = \frac{ab}{a+b+c}.

If segments of lengths p and q emanating from vertex C trisect the hypotenuse into segments of length c/3, then

p^2 + q^2 = 5(c/3)^2.

Bisection

In geometry, bisection is the division of something into two equal or congruent parts, usually by a line, which is then called a bisector. The most often considered types of bisectors are the segment bisector (a line that passes through the midpoint of a given segment) and the angle bisector (a line that passes through the apex of an angle, that divides it into two equal angles).

In three dimensional space, bisection is usually done by a plane, also called the bisector or bisecting plane.

Line segment bisector

A line segment bisector passes through the midpoint of the segment. Particularly important is the perpendicular bisector of a segment, which, according to its name, meets the segment at right angles. The perpendicular bisector of a segment also has the property that each of its points is equidistant from the segment's endpoints. Therefore Voronoi diagram boundaries consist of segments of such lines or planes.

In classical geometry, the bisection is a simple compass and straightedge, whose possibility depends on the ability to draw circles of equal radii and different centers. The segment is bisected by drawing intersecting circles of equal radius, whose centers are the endpoints of the segment. The line determined by the points of intersection is the perpendicular bisector, and crosses our original segment at its center. This construction is in fact used when constructing a line perpendicular to a given line at a given point: drawing an arbitrary circle whose center is that point, it intersects the line in two more points, and the perpendicular to be constructed is the one bisecting the segment defined by these two points.

Angle bisector

An angle bisector divides the angle into two angles with equal measures. An angle only has one bisector. Each point of an angle bisector is equidistant from the sides of the angle.

The interior bisector of an angle is the line or line segment that divides it into two equal angles on the same side as the angle. The exterior bisector of an angle is the line or line segment that divides it into two equal angles on the opposite side as the angle.

To bisect an angle with straightedge and compass, one draws a circle whose center is the vertex. The circle meets the angle at two points: one on each leg. Using each of these points as a center, draw two circles of the same size. The intersection of the circles (two points) determines a line that is the angle bisector.

The proof of the correctness of these two constructions is fairly intuitive, relying on the symmetry of the problem. It is interesting to note that the trisection of an angle (dividing it into three equal parts) cannot be achieved with the ruler and compass alone (this was first proved by Pierre Wantzel).

The angle bisectors of the angles of a triangle are concurrent in a point called the incenter of the triangle.

If the side lengths of a triangle are a,b,c, the semiperimeter ((a+b+c)/2)) is s, and A is the angle opposite side a, the length of the internal bisector of angle A is

2[bcs(s-a)]^{1/2}/(b+c).

If the bisector of angle A in triangle ABC has length t_a and if this bisector divides the side opposite A into segments of lengths m and n, then

t_a^2+mn = bc

where b and c are the side lengths opposite vertices B and C; and the side opposite A is divided in the proportion b:c.

If the bisectors of angles A, B, and C have lengths t_a, t_b, and t_c, then

\frac{(b+c)^2}{bc}t_a^2+ \frac{(c+a)^2}{ca}t_b^2+\frac{(a+b)^2}{ab}t_c^2 = (a+b+c)^2.


From Yahoo Answers

Question:Prove that the bisector of the angle at the apex of an isosceles triangle bisects the base at right angles.

Answers:Given isosceles triangle ABC, where BD bisects ABC. Prove that ADB and CDB = 90 1) ABD = CBD (bisected) 2) AB = CB (isosceles) 3) A = C (isosceles) 4) ABD DAB (ASA) 5) ADB = CDB (congruent triangles) 7) ADC = 180 (straight line) 8) ADB and CDB = 90 each

Question:(the answers are either sometimes, always or never) 1. The circumcenter of a triangle is in the interior of the triangle. 2. The incenter of a triangle is in the exterior of the triangle. 3. The circumcenter of a right triangle is on the hypotenuse of the triangle. 4. The centroid of a triangle is in the exterior of the triangle. 5. The orthocenter of an obtuse triangle is in the interior of the triangle. 6. If G is the centroid of triangle ABC and M is the midpoint of segment BC, then AG to AM is in a ratio of 2:3. 1. The line that passes through the orthocenter, centroid, and circumcenter is called_______ (platos proposition, Euclids line, Eulers line, concurrency line) 2. When a circle is circumscribed about a triangle, the center of the circle is formed by the intersection of the __________________ of the triangle. (perpendicular bisectors, medians, altitudes, angle bisectors) 3. In triangle ABC, D is the centroid and M is the midpoint of segment AC. If BD=6 and DM=2x-5, find MB. (4, 6, 9, none of the above) 4. In acute triangle ABC, segment AD is an altitude, the measure of angle ADC = 2x2+40, BD=3x+8, and DC=4x+4. Find BC. (show all work) 5. Based on the information from the previous problem, explain whether or not triangle ABC is an equilateral triangle. Make sure to support your answer. 6. Given triangle ABC, A(-5,2), B(7,4), C(3,-9). If D(1,3), determine if segment CD is an altitude, median, perpendicular bisector, or angle bisector. Show work to support your answer.

Answers:1 Sometimes 2 Never 3 Always

Question:always, sometimes, or never. at First I put always... but I've drawn pictures and somehow I now think its sometimes.. Am I right?

Answers:I'm pretty sure it's always, no matter what triangle it is the bisector will cut through the centre of the angle and intersect at least one side of the triangle...

Question:Also need to know how to make an angle bisector using only a right triangle tool. I've only done constructions with a compass and my teacher didn't teach me this so I'm confused...I'm not like a slacker I just really don't get it. Any help would be appreciated. :) (Didn't know if I should categorize this in homework or math so I put homework. If anyone's mad I'll delete it)

Answers:You could align the segment horizontally, lay one leg of the right triangle tool above and along the segment so that the right angle vertex is at one endpoint then draw the other leg straight up, then flip it and do the same leg straight down at the other end, then connect the two ends; this will bisect the segment, then make the perpendicular line through that midpoint with the tool. And the angle bisector, put one of the acute angles of the tool at the vertex of the angle to be bisected, line up the leg with one side of the angle, then draw in the other leg of the right triangle (which will be perpendicular to the side of the angle to be bisected; then flip the tool and do the same to the other side of the angle to be bisected; where the two lines you drew cross, that's there the bisector will go through. This would have been way easier to draw rather than type- good luck figuring it out!

From Youtube

Bisectors of the Angles of the Orthic Triangle :demonstrations.wolfram.com The Wolfram Demonstrations Project contains thousands of free interactive visualizations, with new entries added daily. The altitudes of an acute triangle bisect the internal angles of the orthic triangle. Contributed by: Jay Warendorff

Triangle-Angle-Bisector Theorem :Using the Triangle-Angle-Bisector Theorem to solve a problem.