5.3+Medians+and+Altitudes

** 5.3 Medians and Altitudes ** ...working with Laura Bilash...**
 * By Deandra Donato

In this particular section, we will examine the two types of segments of triangles which include medians and altitudes. You will learn various definitions, methods of solving problems, and how to calculate different distances based on your given. These examples will be based on the geometry textbook (pages 279-286), notes, and other useful sites that will be listed. My goal is for you to have an understanding of medians and altitudes as well as being able to figure out example problems, which are given below.

The median is relating to, located in, or extending toward the middle. Also, it is relating to or constituting the middle value in a distribution. In other words, "the median of a triangle is a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side".-Geometry Textbook
 * __Median-__**

Diagram 1.** In this example, segment AD is considered the median since point D is the midpoint.
 * Example:

Diagram 2.
 * There are three medians of a triangle where they are all concurrent. This point of concurrency is known as the centroid of the triangle. In the next three examples, the point of concurrency is point P.

The centroid of a triangle is twice as far from a given vertex than it is from the midpoint to which the median from that vertex goes.

In simpler terms:**  ||  ||||||||  || //BF //, and //CP // = //CE //. ||  || 
 * 
 * THEOREM 5.7 || [[image:http://images.classwell.com/ebooks/images/mcd_geo/blank.gif width="5"]] || //Concurrency of Medians of a Triangle // || [[image:http://images.classwell.com/ebooks/images/mcd_geo/blank.gif width="10"]] ||
 * [[image:http://images.classwell.com/ebooks/images/mcd_geo/blank.gif height="15"]] ||
 * <span style="font-family: 'Comic Sans MS',cursive;">The medians of a triangle intersect at a point that is two thirds of the distance from each vertex to the midpoint of the opposite side.[[image:http://images.classwell.com/ebooks/images/mcd_geo/blank.gif height="8"]]If //<span style="font-family: 'Comic Sans MS',cursive;">P //<span style="font-family: 'Comic Sans MS',cursive;"> is the centroid of Δ //<span style="font-family: 'Comic Sans MS',cursive;">ABC //<span style="font-family: 'Comic Sans MS',cursive;">, then //<span style="font-family: 'Comic Sans MS',cursive;">AP //<span style="font-family: 'Comic Sans MS',cursive;">=[[image:http://images.classwell.com/ebooks/images/mcd_geo/frac_2_3.gif align="absmiddle"]]//<span style="font-family: 'Comic Sans MS',cursive;">AD //<span style="font-family: 'Comic Sans MS',cursive;">, //<span style="font-family: 'Comic Sans MS',cursive;">BP //<span style="font-family: 'Comic Sans MS',cursive;"> =
 * Diagram 3.**

In the previous diagram, diagram 3, P is the centroid. By knowing this information and that PD =4, find AP and AD.
 * Example Problem:**

(pg 279-281 of textbook) <span style="font-family: arial,helvetica,sans-serif;"> (The 2x, x method.)
 * If you do not want to use the 2/3 method as shown above, you can use different method.*

1) Since the smallest side is x, then x=4. The longer side is AP meaning that is 2x. Plug x into 2x (multiply 4X2) and as a result, AP=8. Then, add PD and AP to get a total of 12. Therefore, AD=12.
 * __Given- PD= 4__**

2) If you are given AD = 12, you set up the equation 2x + x = 12. (you will then get x=4 and you can go from there as we did the previous problem)

3) If you are given AP (the longer side) is equal to 8, you divide it by 2, and get PD=4. Then you add and 8+4 and get AD=12.

Now you try!!! -Visit the link below and answer the following questions. Answers are on other wiki page.
 * __Practice Problems:__**

Given: Point L is the centroid. 1) If segment EL is equal to 33, what is the measurement of segment LI and EI? 2) If EJ = 6x-3 and JG = 7x-1, what does x equal? 3) If segment HL is 6, what is the measurement of segment LG? 4) If segment FJ equals 21, what does segment FL and LJ equal?

<span style="font-family: Verdana,Arial,Helvetica,'MS Sans Serif'; font-size: 12px; line-height: normal;">An **altitude of a triangle** is the perpendicular segment from a vertex to the opposite side or to the line that contains the opposite side. An altitude can lie inside, on, or outside the triangle. In other words, it is straight line through a vertex and perpendicular to the opposite side or an extension of the opposite side. Every triangle has three altitudes. The lines containing the altitudes are concurrent and intersect at a point called the **orthocenter of the triangle.** __example of multiple altitudes in a triangle__
 * __Altitude:__**

__More examples:__

<span style="font-family: Verdana,Arial,Helvetica,'MS Sans Serif';"> __**Acute Triangle Right triangle Obtuse Triangle**__ <span style="font-family: Verdana,Arial,Helvetica,'MS Sans Serif'; font-size: 12px; line-height: normal;">Point G is the orthocenter. KM and LM= legs, but The three lines containing The three altitudes intersect are also altitudes. They the altitudes intersect at at the orthocenter at a point intersect at the right angle point W, which is inside the triangle. meaning the orthocenter is outside the triangle. located on the triangle.

(pg 281 of textbook)
 * [[image:http://images.classwell.com/ebooks/images/mcd_geo/blank.gif width="1"]] || [[image:http://images.classwell.com/ebooks/images/mcd_geo/blank.gif width="10"]] || THEOREM || [[image:http://images.classwell.com/ebooks/images/mcd_geo/blank.gif width="10"]] || [[image:http://images.classwell.com/ebooks/images/mcd_geo/blank.gif width="1"]] ||
 * [[image:http://images.classwell.com/ebooks/images/mcd_geo/blank.gif width="1"]] || [[image:http://images.classwell.com/ebooks/images/mcd_geo/blank.gif width="10"]] ||  |||||||| [[image:http://images.classwell.com/ebooks/images/mcd_geo/blank.gif height="15"]] ||
 * **THEOREM 5.8** || [[image:http://images.classwell.com/ebooks/images/mcd_geo/blank.gif width="5"]] || //Concurrency of Altitudes of a Triangle// || [[image:http://images.classwell.com/ebooks/images/mcd_geo/blank.gif width="10"]] ||
 * [[image:http://images.classwell.com/ebooks/images/mcd_geo/blank.gif width="461" height="5"]] ||
 * The lines containing the altitudes of a triangle are concurrent.[[image:http://images.classwell.com/ebooks/images/mcd_geo/blank.gif height="8"]][[image:http://images.classwell.com/ebooks/images/mcd_geo/blank.gif height="8"]]If [[image:http://images.classwell.com/ebooks/images/mcd_geo/linesegment_ae.gif align="absmiddle"]], [[image:http://images.classwell.com/ebooks/images/mcd_geo/linesegment_bf.gif align="absmiddle"]], and [[image:http://images.classwell.com/ebooks/images/mcd_geo/linesegment_cd.gif align="absmiddle"]] are the altitudes of Δ//ABC//, then the lines [[image:http://images.classwell.com/ebooks/images/mcd_geo/line_ae.gif align="absmiddle"]], [[image:http://images.classwell.com/ebooks/images/mcd_geo/line_bf.gif align="absmiddle"]], and[[image:http://images.classwell.com/ebooks/images/mcd_geo/line_cd.gif align="absmiddle"]] intersect at some point //H//. ||
 * [[image:http://images.classwell.com/ebooks/images/mcd_geo/mcd_ma_geo_lsn_0395937779_p279_f14.gif width="160" height="113" align="top"]] ||  ||   || [[image:http://images.classwell.com/ebooks/images/mcd_geo/blank.gif width="1"]] ||
 * [[image:http://images.classwell.com/ebooks/images/mcd_geo/blank.gif height="1"]] ||

In terms of our triangle, this theorem simply states what we have already shown: since AD is the altitude drawn from the right angle of our right triangle to its hypotenuse, and CD and DB are the two segments of the hypotenuse. http://jwilson.coe.uga.edu/EMT668/EMAT6680.Folders/Brooks/6690stuff/Righttriangle/rightday3.html
 * Right Triangle Altitude Theorem Part a: **The measure of the **altitude** drawn f**rom the vertex of the right angle** of a right triangle to its hypotenuse is the**geometric mean between the measures of the two segments of the hypotenuse**.

-Now you try!!! Visit the link below and answer the following questions. Answers are on other wiki page.
 * __Practice Problems:__**

Given: Segment BD is an altitude. 1) If segment AD = 2x, segment DC = 3x-4 and the right angle (angle BDC) =4x+10, what is x? 2) Using the above problem, what is segment AC? 3) Using question 1, what is segment AD? 4) Using question 1, what is segment DC?

__//True or False?//__
 * Overall questions:**

1) The median of a triangle could also be the perpendicular bisector. 2) The altitude of a triangle could also be the perpendicular bisector. 3) The medians of triangles always intersect inside the triangle. 4) The altitudes of a triangle always intersect inside the triangle.

For additional information on altitudes and terminology, visit this link and feel free to visit the links noted before. http://www.mathopenref.com/triangle.html