Geometry: Medians and Midsegment
This is a free lesson from our course in Geometry
 
   
This lesson explains the important theorems in relation to medians and midsegment . You’ll, learn it heading further to the earlier learning on basic properties of altitudes, medians, midsegments, angle bisectors, and perpendicular bisectors of triangles. The contents and explanation with examples are presented by instructor using video and in own hand writing. This will provide the geometry help; in context with triangles shapes, figures and applying the developed skills, for finding solution to real-world problems.
Theorems for segments in Triangles
The lines or line segments like- altitudes, medians, midsegments, angle bisectors, and perpendicular bisectors of triangles, within triangles are concurrent i.e. all of them share intersecting points. The intersecting point is called the point of concurrency, and this has special properties for such lines or line segments. (More text below video...)
<h2> Medians and Midsegment </h2> <p> median, triangle, video, line, measure, medians and midsegments, ratio, point, segment, midpoint, equal, geometry help, parallel, vertex, opposite, side, centroid, solution, coordinate, midsegment, geometry tutorials, quizzes</p> <p> The medians of a triangle are the segments drawn from the vertices to the midpoints of the opposite sides and divide them in the ratio of 2:1. </p>
Other useful lessons:
Triangles Classification
Triangle Angle-Sum Theorem
Triangle Inequality
(Continued from above) Altitudes: The lines containing the altitudes of a triangle meet at a point known as orthocenter of the triangle. Notice that the orthocenter may not be necessarily in the interior of the triangle.
The other important theorems in relation to altitudes of a triangle consider the concept of similarity i.e.
• the lengths of the altitudes of similar triangles have the same proportions as the corresponding sides of the similar triangles.
• the altitude of a right angle triangle drawn from the vertex making right angle at the hypotenuse divides the triangle into two similar triangles. These triangles are also similar to the original triangle. E.g. PQR, ARP and AQP are similar triangles (symbolically represented as) i.e. PQR  ~ ARP ~ AQP (Fig-2, above).
Medians: The medians of a triangle are segments from each vertex to the midpoint of the opposite side. A triangle has three medians and these medians intersect at a point and this point is called the ‘centroid’ of the triangle. The ‘centroid’ divides each median in the ratio 2:1, starting from the vertex. For example: (in the figure below),
if a median is drawn from vertex A to midpoint D through centroid O, the length of AO is twice the length of OD i.e. the centroid is 2/3 away from the vertex to the opposite side midpoint. E.g. If segment AD & BE are the medians of  BAC, the length of segment AD is 12 units, if AO is 8 units long. Note that the centroid is always on the interior of the triangle.
It is important to remember:
• the lengths of the medians of similar triangles are of the same proportion as the lengths of corresponding sides.
• the length of median of a right triangle, from the right angle to the hypotenuse is half of the hypotenuse.
Medians Statement
The medians of a triangle pass through the same point and that divides each of the medians in the ratio 2:1.
Given: In BAC, medians AD and CF intersect at O. BG is joined to meet AC at E.
To prove: BE is the median i.e. AE = EC, and
               BO: OE = AO: OD = CO: OF = 2:1
Required Construction: Produce BE to G such that BO = OG. Then join GA and GC.
Proof:
(1) BE is the median i.e. AE = EC
In BAG, F is the mid-point of BA (given), and O is the mid-point of BG (by construction). Therefore, FO || AG or OC || AG --- (a)
By theorem- line joining the mid-points of any two sides of a triangle.
Similarly in BCG, OD || CG, or OA || GC and OD = ½ CG. ---- (b)
From (a) and (b) above, AGCO is a parallelogram.
Now in AGCO, AC & OG are diagonals and they bisect each other. Therefore, AE = EC or BE is the median from vertex B.
(2) OG and AC are diagonals of OCGA.
OE = EG
OE = ½ OG
OE = ½ BO or BO/OE = 2:1, i.e. BO: OE = 2:1
Having similar approach; if you work out for other two, AO: OD and CO: OF will have the ratio of 2:1. Hence it is proved.
Median Length: length of a median can be found, if lengths of the sides of a triangle are known. E.g. say three original sides of the triangle are a, b, and c. Then length of the median ‘ma’ from vertex A to side a is given by, 
ma =
The second relationship is – the sum of the squares of the medians is ¾, the sum of the squares of the sides, i.e.
ma2+ mb2+ mc2= (3/4) (a2+b2+c2).
Midsegments of a Triangle:
It is defined as the segment whose endpoints are both midpoints of sides. Every triangle has three midsegments.
Triangle Mid-segment Theorem
Segment connecting the midpoints of any two sides of a triangle is parallel to the third side i.e. whose end points are both midpoints of sides, and is half as long (see figure below). For example,
Given: ABC with point midpoint of AC, point the midpoint of BC and point is the midpoint of AB, we can conclude as follows:
                                          || AB                      
                                          = ½ AB
                                          || AC                      
                                          = ½ AC                      
                                          || BC                      
                                          = ½ BC                        
It may be seen that you’ll end up with four triangles that are congruent.
Angle Bisectors of Triangles:                                         
The angle bisectors of a triangle intersect each other at a point. This point is called the incircle of the triangle. The incircle of a triangle and center of a circle inscribed in a triangle means the same. The important property to note here is that the incircle is equidistant from the three sides of the triangle. It is based on the logic of inherent congruency of the radii of a circle. Other property relates to the angle bisectors dividing the opposite side in the ratio of the adjacent sides. For example:
Angle bisectors divide the opposite side in the ratio of the adjacent sides. E.g. in

ABC, AP is an angle bisector of A, then
                                        AB/AC = PB/PC
Notice that the same is true for the external bisectors also. 
Proof:
Draw the straight line through C parallel to AP, such that it meets AB in E. Then, note thatAEC is isosceles: AC = AE. This is because
             ACE    = CAP    = BAP    = AEC.
Therefore, AE = AC, and the required proportion follows from the similarity of BEC and BAP.
This also shows that the three angles bisectors in a triangle meet in a point. As there is only one point on a given segment that divides it in a given ratio, converse theorem holds true. Thus if a point divides the base of a triangle in the ratio equal to the ratio of the sides, it shall be the origin of the angle bisector from the apex.
Perpendicular Bisectors:
As learnt earlier perpendicular bisector of a triangle is a straight line passing through the midpoint of a side and being perpendicular to it, i.e. it results in a right angle with it. The important property to notice is that the three perpendicular bisectors meet at a single point, known as cirumcenter of the triangle that is equidistant from the vertices of the triangle. This point is the center of the circumcircle, which passes through the three vertices of the triangle. The radius of circumcircle can be determined using the law of sines.
The position of circumcenter ‘O’ depends on the type of triangle (see figure below):
• if and only if a triangle is acute (angles are smaller than a right angle), the circumcenter lies inside the triangle.
• if and only if it is obtuse (one angle larger than a right angle), the circumcenter lies outside.
• if and only if it is a right triangle, the circumcenter lies on the hypotenuse.
Remember: the circumcenter of a triangle may not necessarily exist in the interior of the triangle and its position depends on the type of triangle as seen in the above figure.
The video above will explain more details about the Altitudes, Medians and Midsegment and relationship of the involved properties, with the help of several examples and practice problems.
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