Relation between Median and Sides of a Triangle

Introduction to Median and Sides of a Triangle

A triangle has 3 medians. A line segment that joins any vertex of the triangle and the mid-point of its opposite side is called a median. It is also the line from the midpoint of a side to the opposite interior angle. They are concurrent at the centroid. The point which is common to all the 3 medians during their crossing is called a point of concurrency, the centroid of a triangle.

How to Find the Length of the Median?

The different ways to find the length of a median are as follows:

• The formula for the length of the median to side
  \(\begin{array}{l}BC = \frac{1}{2}\sqrt{2AB^{2}+2AC^{2}-BC^{2}}\end{array} \)

• Using Apollonius’s Theorem, the formula for the median of a triangle is given by

Properties of Median of a Triangle

The properties of the median are as follows:

• The median bisects the vertex angle in an isosceles and equilateral triangle where the two adjacent sides are the same.
• The three medians of a triangle intersect at a point called the centroid.
• The area of the triangle is divided into half by a median.
• The triangle is divided into 6 smaller triangles of the same area by the centroid.
• In an equilateral triangle, the length of the medians is equal.
• The medians from the vertices having equal angles are of equal length in the case of an isosceles triangle.
• The length of the medians differs in a scalene triangle.
• The sum of two sides of a triangle is greater than the median drawn from the vertex, which is common.
• The median and lengths of sides are related in such a way that “3 times the sum of squares of the length of sides = 4 times the squares of medians of a triangle.”

3 (AB² + BC² + CA²) = 4 (AD² + BE² + CF²).

• All three medians intersect at one single point that divides the medians’ lengths in the ratio of 2:1.

Solved Problems on Relation between Median and Sides of a Triangle

Example 1: In a ΔABC, AD, BE and CF are three medians. Then the ratio

Solution:

In any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median bisecting it.

∴ AB² + AC² = 2 (AD² + BD²)

⇒ AB² + AC² = 2 (AD² + BC² / 4)

⇒ 2(AB² + AC² ) = 4 AD² + BC²

Similarly, ⇒ 2 (AB² + BC²) = 4 BE² + AC²

⇒ 2(AC² + BC²) = 4 CF² + AB²

On adding all three 4 (AB² + BC² + AC²) = 4 (AD² + BE² + CF²) + BC² + AC² + AB²

⇒ 3 (AB² + BC² + AC²) = 4 (AD² + BE² + CF²)

⇒ [AD² + BE² + CF²] / [AB² + BC² + AC²]

= 3 / 4

Example 2: In the adjoining figure given, ∠PQR = 90^∘ and QL is a median, PQ = 12 cm, and QR = 16 cm. Then, QL is equal to __________.

Solution:

Given that, PQ = 12 cm, QR = 16 cm and QL is a median.

∴ PL = LR  …….(I)

In ΔPQR, (PR)² = (PQ)² + (QR)² (By Pythagoras theorem)

= 144 + 256 = 400

⇒ PR = 20

Now, by Pythagoras theorem, if L is the mid-point of the hypotenuse PR of a right-angled ΔPQR, then

QL = 1 / 2 [PR] = [1 / 2] * (20) = 10 cm

Example 3: In a triangle, ABC, coordinates of A are (1, 2), and the equations of the medians through B and C are x + y = 5 and x = 4, respectively. Then, what is the area of ΔABC (in sq. units)?

Solution:

Median through C is x = 4

So, clearly, the x coordinate of C is 4.

So, let C = (4, y), then the midpoint of A (1, 2) and C (2, y), which is D, lie on the median through B by definition.

Clearly, D = ([1 + 4] / 2, [2 + y] / 2).

Now, we have, [3 + 4 + y] / 2 = 5 ⇒ y = 3. So, C = (4, 3).

The centroid of the triangle is the intersection of the medians.

It is easy to see that the medians x = 4 and x + y = 5 intersect at G = (4, 1).

The area of triangle ΔABC = 3 × ΔAGC = 3 ×(1 / 2) × 3 × 2 = 9.

(In this case, it is easy as the points G and C lie on the same vertical line x = 4. So, the base GC = 2 and the altitude from A is 3 units.)

Example 4: If the median of ΔABC through A is perpendicular to AB, then find the relation between angles A and B.

Solution:

We have BD = DC and ∠DAB = 90^o.

Draw CN perpendicular to BA produced, then in ΔBCN, we have DA = 1 / 2 CN and AB = AN

Let ∠CAN = α [∵tanA = tan (π − α) = −tanα = −CN / NA =−2 [AD / AB] = −2 tanB

⇒ tan A + 2 tanB = 0.

Example 5: If a triangle PQR, sin P, sin Q, sin R are in A.P., then the altitudes are in _________.

Solution:

Let p₁, p₂, and p₃ be altitudes from P, Q, and R.

By sine rule, a/sin P = b/sin Q = c/sin R = k

Also ½ ap₁ = ∆

p₁ = 2∆/k sin P

Similarly, p₂ = 2∆/k sin Q

p₃ = 2∆/k sin R

Given, sin P, sin Q, sin R are in AP.

So, 1/sin P, 1/sin Q, 1/sin R are in HP.

2∆/k sin P, 2∆/k sin Q, 2∆/k sin R are in HP.

=> p₁, p₂, p₃ are in HP.

Example 6: In the figure given below, two medians AD and BE of ΔABC intersect at G at right angles. If AD = 18cm and BE = 12cm, then the length of BD (in cm) is _________.

Solution:

AD = 18 cm ⇒ GD = [1 / 3] × 18 = 6 cm

BE = 12 cm ⇒ BG = [2 / 3] × 12 = 8 cm

∴ BD = √[6² + 8²] = 10 cm

Example 7: Let PS be the median of the triangle with vertices P (2, 2), Q (6, − 1)and R (7, 3). The equation of the line passing through (1, -1) and parallel to PS is _____________.

Solution:

S = midpoint of QR = ([6 + 7] / [2], [−1 + 3] / 2) = (13 / 2, 1)

The slope of PS = [2 − 1] / [2 − 13 / 2] = −2 / 9,

The required equation is y + 1 = (−2 / 9 )(x − 1)

i.e., 2x + 9y + 7 = 0