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Medians and Altitudes

Sep 12, 2022
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Key Concepts

  • Median and centroid.
  • Altitudes and orthocenter.
  • Special case of isosceles triangle.

Median and Centroid 

Median

A segment from a vertex to the midpoint of the opposite side is called the median of a triangle.  

A triangle has three vertices and three medians. 

The three medians of a triangle are concurrent. 

Centroid

The point of intersection of medians is called the centroid. 

At a particular point, the triangle will balance, this point is called the balancing point. 

parallel

This balancing point is also called the centroid. 

Concurrency of Medians of a Triangle 

Theorem

The medians of a triangle intersect at a point that is two-thirds the distance from each vertex to the midpoint of the opposite side. 

The medians of ∆ABC intersect at point O, then CO =  CD,

AO = 2/3 AE, BO = 2/3 BF

Concurrency of Medians of a Triangle

Note: 

parallel

The medians of a triangle intersect at a point, at which the median is divided in a 2: 1 ratio from the vertex to the midpoint of the opposite side.   

Example 1: 

If in triangle ABC, G is the centroid, if AG = 11 units, find GD, AD. 

Example 1: 

Solution: 

AG = 2/3 AD 

11 = 2/3 AD  

AD = 33/2

AD = 16.5

GD = AD – AG

= 16.5 – 11

= 5.5 units

Therefore, AD = 16.5 units, GD = 5.5 units

Example 2: 

The vertices of ∆ABC are A (1, 3), B (3, 5), and C (5, 4). Which ordered pair gives the coordinates of the centroid G of ∆ABC? 

Example 2: 

Solution: 

The vertices of ∆ABC are A (1, 3), B (3, 5), and C (5, 4) 

Draw the graph and plot the midpoint of AC as D 

The mid-point of AC =( 1+5/2 , 3+4/2 )

D=(3,3.5)

The centroid is at a two-third distance from each vertex to the midpoint of the opposite side.  

The distance from vertex B (3, 5) to D (3, 3.5) is 5 – 3.5 = 1.5 units.  

centroid = 2/3 * 1.5=1 

The coordinates of the centroid G are (3, 5 – 1), or (3, 4). 

Activity

How to find a balance point (median) of a triangle? 

Balancing point: 

Step 1: 

Take a cardboard and cut it in the shape of a triangle.  

Step 2: 

Balance the triangle on the eraser end of a pencil. 

Step 3: 

Mark the point as a balancing point with the marker as B. 

Median: 

Step 1: 

Draw the triangle on the triangle cut out. 

Step 2: 

Using a ruler, find the midpoints of the triangle. 

Step 3: 

Draw a segment from the midpoint to the opposite vertex, and point at the intersection point as M. 

What do you observe by B and M? 

Altitudes and Orthocenter 

Altitudes 

The altitude of a triangle is the segment that is perpendicular to the opposite side from the vertex  or to the line which has an opposite side. 

Altitudes 

Concurrency of Altitudes in a Triangle 

Theorem

The lines that have the altitudes of a triangle are concurrent. 

AF, BD, and CE are concurrent at point O. 

Concurrency of Altitudes in a Triangle 

Orthocenter 

The point at which the lines containing three altitudes of the triangle meet is called the orthocenter or the point of concurrency of the altitudes.  

Example: 

Orthocenter of a triangle 

The orthocenter is the point where all three altitudes of the triangle intersect 

Orthocenter of a triangle 

Special Case 

Isosceles Triangle

In an isosceles triangle, angle bisector, perpendicular bisector, median, and altitude from vertex point to the opposite side, all the segments are the same. 

Remark: 

This is true even in an equilateral triangle with special segments. 

Property of Isosceles Triangle

Property

In an isosceles triangle, prove that the median to the base is an altitude. 

Property of Isosceles Triangle

Proof: 

Given that ∆ABC is an isosceles triangle, AD is median to the base BC. 

Now, we need to prove AD is the altitude to ∆ABC  

AC and AB are isosceles in ∆ABC are congruent. 

CD = BD, as AD is median 

Also, BD = BD 

Now ∆ADC ≅ ∆ABD by SSS congruence postulate, 

∠ADC ≅ ∠ADB are also congruent angles, as there are in congruent triangles. 

∠ADC and ∠ADB are linear pairs of angles, AD intersects BC to form a linear pair of angles. 

So, AD ⊥ BC, AD is the altitude of ∆ABC. 

Hence proved. 

Activity

Let us do this activity to understand the difference between perpendicular bisectors, angular bisectors, medians, and altitudes. 

Step 1: 

Take a cardboard and cut four similar triangles from it. 

Step 2: 

In the first triangle, draw the perpendicular bisectors and find the point of concurrency.  

In the second triangle, draw the angular bisectors and find the point of concurrency, 

In the third triangle, draw the medians and find the point of concurrency 

In the fourth one, draw the altitudes and find the point of concurrency. 

Step 3: 

Now compare the four points of concurrency and understand. 

Activity

If AD = 21 cm, Find AG.

If GC= 16 cm, Find CF.

Solution:

AD = 21 cm

AG = 2/3AD 

 = 2/3 * 21

= 14 cm

GC = 16 cm

CG = 2/3 * CF

16 = 2/3 * CF

CF=24 cm

Exercise

  • Compare a perpendicular bisector and altitude.
  • Compare an angle bisector and median.

In the given figure, G is the centroid

In the given figure, G is the centroid
  • Find GE.
  • Find AB
  • Find BE.
  • Find BF.
  • In the given figure, if GE = 4 units, AD = 12 units, CF = 14 units. Find BG.
In the given figure, if GE = 4 units, AD = 12 units, CF = 14 units. Find BG.
  • Find ∆ABD ≡ ∆ACD
Find ∆ABD ≡ ∆ACD
  • In an isosceles triangle BAC find DC.
In an isosceles triangle BAC find DC.
  • Find ∠BAD.

Concept Map

Concept Map

What have we learned

  • Median and Centroid.
  • Altitude and Orthocenter.
  • Special case in Isosceles Triangle.

Comments:

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