Lesson The length of a median of a triangle

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The length of a median of a triangle


This lesson is focused on the formula expressing the length of a median of a triangle.
The proof of the formula is presented, which is based on the Law of cosines (see the lesson Proof of the Law of Cosines revisited under the topic Trigonometry of
the section Algebra-II in this site). The examples are provided too, showing how to use this formula.

Theorem 1
In a triangle with the sides a, b and c the median drawn to the side c has the length of

m%5Bc%5D = sqrt%282a%5E2%2B2b%5E2-c%5E2%29%2F2. Prove.

Proof
Figure 1 shows the triangle ABC with the sides a (BC), b (AC) and c (AB) and                     
the median m (CD) drawn to the side c. Let d be the length of the segment AD
and e be the length of the segment DB the median CD divides the side AB. Denote
as delta and phi the angles at the intersection the median CD and the side AB.

Apply the Law of Cosines to express the length of the side AC of the triangle ADC
b%5E2 = d%5E2+%2B+m%5E2+-+2dm%2Acos%28delta%29.
Apply the Law of Cosines to express the length of the side BC of the triangle BDC
a%5E2 = e%5E2+%2B+m%5E2+-+2em%2Acos%28phi%29.


    Figure 1. To the Theorem 1
Now, take the sum of the last two equalities. Note that cos%28delta%29+=+-+cos%28phi%29, because the angles delta and phi are supplementary angles. Also note that d+=+e, because CD is
the median. Therefore, when taking the sum, the terms 2dm%2Acos%28delta%29 and 2em%2Acos%28phi%29 cancel each other out. So, after summing we get
b%5E2 + a%5E2 = d%5E2 + e%5E2 + 2m%5E2.

Substitute here d+=+c%2F2 and e+=+c%2F2 using the fact that CD is the median. We get
b%5E2 + a%5E2 = c%5E2%2F4 + c%5E2%2F4 + 2m%5E2,
or
m%5E2 = (2a%5E2 + 2b%5E2 - c%5E2)/4.
This is what to be demonstrated. The proof is completed.

Theorem 2
In a triangle with the sides a, b, and c the lengths of the medians m%5Ba%5D, m%5Bb%5D and m%5Bc%5D, drawn to the sides a, b and c respectively satisfy to the identity
m%5Ba%5D%5E2+%2B+m%5Bb%5D%5E2+%2B+m%5Bc%5D%5E2 = %283a%5E2%2B3b%5E2%2B3c%5E2%29%2F4. Prove.

Proof
Based on the Theorem 1 we have three equalities

m%5Ba%5D%5E2 = %282b%5E2%2B2c%5E2-a%5E2%29%2F4,

m%5Bb%5D%5E2 = %282a%5E2%2B2c%5E2-b%5E2%29%2F4, and

m%5Bc%5D%5E2 = %282a%5E2%2B2b%5E2-c%5E2%29%2F4.

By taking the sum of these equalities, we get the identity

m%5Ba%5D%5E2+%2B+m%5Bb%5D%5E2+%2B+m%5Bc%5D%5E2 = %283a%5E2%2B3b%5E2%2B3c%5E2%29%2F4.

This is exactly what to be demonstrated. The Theorem 2 is proved.


Example 1
In the triangle the side lengths are a = 5, b = 6 and c = 4 (Figure 2).            
Find the length of the median m%5Bc%5D drawn to the side c.

Solution
Apply the formula for the median length

m%5Bc%5D = sqrt%282a%5E2%2B2b%5E2-c%5E2%29%2F2 = sqrt%282%2A5%5E2%2B2%2A6%5E2-4%5E2%29%2F2 = sqrt%282%2A25%2B2%2A36-16%29%2F2 =

            sqrt%2850%2B72-16%29%2F2 = sqrt%28106%29%2F2.

Answer. m%5Bc%5D = sqrt%28106%29%2F2 =~ 5.148 (approximately).


    Figure 2. To the Example 1


Example 2
In the isosceles triangle the lateral side has the length of 4.                             
The median drawn to the lateral side has the length of 3 (Figure 3).
Find the length of the base of the triangle.

Solution
Let x be the length of the base of the triangle.
Apply the formula for the median length. You get the equation

3+=+%281%2F2%29%2Asqrt%282%2A4%5E2+%2B+2%2Ax%5E2+-+4%5E2%29.

Simplify this equation step by step as shown below:

9+=+%281%2F4%29%2A%282%2A16+%2B+2%2Ax%5E2+-+16%29,

9+=+%281%2F4%29%2A%2816+%2B+2%2Ax%5E2%29,

36+=+16+%2B+2%2Ax%5E2,

x%5E2+=+36-16,

x%5E2+=+10,

x+=+sqrt%2810%29.

Answer. The base of the triangle has the length of sqrt%2810%29.


    Figure 3. To the Example 2


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