Module 1

Chapter I
Fundamental Concepts
1.1. The Cartesian Coordinate System
1.2. Distance Between Two Points
1.3. Inclination And Slope Of A Line
1.4 Angle Between Two Lines
1.5. Division Of A Line Segment
1.6. Analytic Proofs Of Geometric Theorems

1
 Objectives:

At the end of the chapter, the student should be able to:

1. define a line segment;
2. identify the coordinate of points – abscissa and ordinate;
3. plot the points on a coordinate axes;
4. determine the quadrant where point terminates;
5. compute the distance between two points;
6. determine the coordinates of the midpoint of a line;
7. compute the slope of a line;
8. identify lines as parallel or perpendicular base on their slopes;
9. determine the measure of angles formed by two intersecting
lines;
10. determine the measure of angles formed by two intersecting
lines;
11. determine the coordinate of a point in the division of a line
segment;
12. determine the area of a triangle given its three vertices; and
13. give an analytic proofs to geometric theorems.

2
 Learning Activities/Content

1.1. THE CARTESIAN COORDINATE SYSTEM

This section aims to:

1. define a line segment;
2. identify the coordinate of points – abscissa and ordinate;
3. plot the points on a coordinate axes; and
4. determine the quadrant where point terminates.

Directed Line Segment

A line segment is a set of points consisting of two distinct points

together with all points between them.

line segment P1P2

P1 P2

When a line is measured in a definite sense from one endpoint to the

other, the segment is said to be directed. If the terminal points P1 and P2,
in the line segment P1P2 or in the segment P2P1, then the sense is from P1
to P2 or from P2 to P1. Thus, if P1P2 is considered positive, then P2P1 is
treated negative, that is P1P2 = -P2P1.

If P3 is any third point of the same straight line through P1 and P2, i.e.

P1 P2 P3 P1 P3 P2 P3 P1 P2

then for all possible positions of P1, P2 and P3, P1P3 = P1P2 + P2P3.

Examples:

1.
A B C
In the above figure, the positive direction is to the right.
Thus, AB=3, BC=2, AC=5, and CA=-5

3
 2.
B A C

In the above figure, the positive direction is also to the right.

Thus, AB=-1, BC = 5, AC = 4, and CA= -4

Rectangular Coordinate System

The Rectangular Coordinate Plane also known as Cartesian

Coordinate Plane is formed by two perpendicular lines where their point
of intersection is called the origin, usually denoted by O. The horizontal
line is called the horizontal axis or x-axis with positive numbers located
to the right of the origin and the vertical line is called the vertical axis or
y-axis with the positive numbers lying above the origin. The axes divide
the plane into four quadrants which are named from I to IV in a counter
clockwise direction, beginning from the upper right portion.
The plane has infinite points. Each point P corresponds to a unique
ordered pair (x, y), x is called the abscissa which tells the distance of
point P from the vertical axis and y is called the ordinate which tells the
distance of point P from the horizontal axis.
The signs of the coordinates of points in the four quadrants are
shown in the figure below.

y y

a
b P (a, b)
QII (-, +) QI (+, +)
b
x
O
QIII (-, -) QIV (+, -) x
a

4
Examples: Plot the points A(-5, 1), B(4, 3), C(-1, -2) and D(2, -5) on
the Cartesian Coordinate Plane and determine the quadrant
where it lies.

Solution:
y

A
x

Point A with coordinate (-5, 1) is located 5 units to the left of

the y-axis and 1 unit above the x-axis. Thus, the abscissa is
-5 and its ordinate is 1. Point A terminates in the 2nd
quadrant.

Point B with coordinate (4, 3) is located 4 units to the right

of the y-axis and 3 units above the x-axis. Thus, the abscissa
is 4 and its ordinate is 3. Point B terminates in the 1st
quadrant.

Point C with coordinate (-1, -2) is located 1 unit to the left of

the y-axis and 2 units below the x-axis. Thus, the abscissa is
-1 and its ordinate is -2. Point C terminates in the 3rd
quadrant.

Point D with coordinate (2, -5) is located 2 units to the right

of the y-axis and 5 units below the x-axis. Thus, the abscissa
is 2 and its ordinate is -5. Point D terminates in the 4th
quadrant.

5
6
 EXERCISES 1.1

Name: _____________________________________________________________ Score: ______________

Section: ___________________________________________________________ Date: ____________________________

Plot the following points on a Cartesian coordinate plane below and state the
quadrant in which it lies.

Quadrant Quadrant
1. A(-1, 2) ________ 6. O(-6, -7) ________
2. N(2, -3) ________ 7. M(-7, 8) ________
3. L(-3, -4) ________ 8. T(8, -9) ________
4. G(-4, 5) ________ 9. R(-9, 0) ________
5. E(5, 6) ________ 10. Y(0, 1) ________

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8
1.2. DISTANCE BETWEEN TWO POINTS

This section aims to:

1. compute the distance between two points; and
2. determine the coordinates of the midpoint of a line.

Recall: Absolute Value

The absolute value of x, denoted by lx l, is the distance from zero

on the number line.

x, if x is positive
lx l = 0, if x = 0
-x, if x is negative

Examples: /10/ = 10; /0/ = 0; /-1 / = -(-1) = 1

Distance Between Two Points

The distance between two points P1(x1, y1) and P2(x2, y2) is the
number of units measured along the line between the two points.

y
If the line is parallel to the
x-axis, then y1 = y2, and the
distance is given by
P1(x1, y1=y2) P2(x2, y2=y1)

d = /P1P2/ = /x2 – x1/ = /x1 – x2/ x

Example 1: Find the distance between the points (1, 0) and (9, 0)

Solution: d = /P1P2/ = /x2 – x1/ = /9 – 1/ = /8/ = 8

Example 2: Find the distance between the points (-1, 0) and (1, 0)

Solution: d = /P1P2/ = /x2 – x1/ = /1 – (-1)/ = /2/ = 2

9
 y
If the line is parallel to the
P1(x1=x2, y1)
y-axis, then x1 = x2, and the
distance is given by x

d = /P1P2/ = /y2 – y1/ = /y1 – y2/ P2(x2=x1, y2)

Example 1: Find the distance between the points (0, 12) and (0, -2)

Solution: d = /P1P2/ = /y2 – y1/ = /-2 – 12/ = /-14/ = 14

Example 2: Find the distance between the points (0, -4) and (0, -9)

Solution: d = /P1P2/ =/y2 – y1/ = /-9 – (-4)/ = /-5/ = 5

y
If a line is not parallel to either
axes, then the distance can be found P1(x1, y1)

by using Pythagorean Theorem.

Theorem. The distance between any two points

P1(x1, y1) and P2(x2, y2) is given by the P2(x2, y2)
formula x

d = /P1P2/ = √ (x2 – x1)2 + (y2 – y1)2

Example 1: Find the distance between the points (2, 0) and (-2, 3)

Solution: d = √ (x2 – x1)2 + ( y2 – y1)2

= √ (-2 – 2)2 + (3 – 0)2
= √ (-4)2 + 32
= √ 16 + 9
= √ 25
= 5

10
 Example 2: Find the distance between the points (-3, 1) and (9, 6)

Solution: d = √ (x2 – x1)2 + ( y2 – y1)2

= √ [(9 – (-3)]2 + (6 –1)2
= √ (12)2 + (5)2
= √ 144 + 25
= √ 169
= 13

Midpoint of a Line Segment

y
In figure at the right,
P2
let P(x, y) be the midpoint of the (𝑥 , 𝑦
P(x,
segment joining the points P1(x1, y1)
P1 (𝑥 , 𝑦 y)
and P2(x2, y2).

Thus, OR = OR1 + R2
x = OR1 + (OR2 – OR1)
= x1 + (x2 – x1) x
O R1(𝑥 , R(𝑥 ,
x = (x1 + x2) RR22(𝑥 ,

In getting for the value of y, similar procedure shall be followed.

Thus, y = (y1 + y2).

Hence, the coordinates of the point P(x, y) is midway between

points P1(x1, y1) and P2(x2, y2), that is

P(x, y) = P ( ,

Example 1: Find the midpoint between the points P1(2, -5) and
P2(-2, -1).

Solution: ( ,
( (
( ,

(,
( ,

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 Example 2: Find the midpoint between the points P(7, 2) and
Q(-6, -1).

Solution: ( ,
( (
( ,
(½, ½)

Example 3: Show that the quadrilateral with vertices P1 ( , 4),

P2 ( , 3), P3 (1, 0), P4 (3, 1) is a parallelogram.

Solution: Note that a parallelogram is a quadrilateral where its

diagonals bisect each other. Thus, the diagonals have
the same midpoint.

P1
P2

P4
x
P3

For the midpoint of P1P3: For the midpoint of P2P4:

( , ( ,

( , ( ,

( , ( ,

Since the midpoints of diagonals coincide, then

the diagonals bisect each other, hence the
quadrilateral is a parallelogram.

12
 EXERCISES 1.2

Name: _____________________________________________________________ Score: ______________

Section: ___________________________________________________________ Date: ____________________________

I. Find the distances between the given pairs of points

1. (2,7); (-1, 4)

2. (-3, 4); (2, -8)

3. (5, -12); (0, 0)

II. Draw the triangle with vertices P(1, -1), Q(4, -1), R(4, -3), then find the
length of each sides.

y
x

13
 III. Determine whether the points X(-5, 7), Y(2, 6) and Z(1, -1) are all equidistant
from W(-2, 3).

IV. Show that the points H(-1, -2), J(2, 1), L(-3, 6) are vertices of a right triangle.

V. Show that the points L(4, 5), M(1, 1) and N(-2, 5) are vertices of an isosceles
triangle.

14
 VI. If point D(x, 3) is equidistant from E(3, -2) and F(7, 4), find x.

VII. Find the coordinates of the midpoint of the line segment joining each of the
following pairs of points:

1. (9, 1) and (6, 4)

2. (2, 3) and (-6, -5)

3. (-3 ,4) and (3, -4)

VIII. If the line segment connecting P1(x1, 7) and P2(10, y2) is bisected by the point
P(8, 2), find the values x1 and y2.

15
 IX. Find the coordinates of the midpoints and lengths of the median, that is the
line segment joining the vertex of a triangle and the midpoint of the opposite
side, of the triangle with vertices at points A(1, 5), B(5, -1) and C(11, 9).

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1.3. INCLINATION AND SLOPE OF A LINE

This section aims to:

1. compute the slope of a line; and
2. identify lines as parallel or perpendicular base on their slopes; and
3. determine the measure of angles formed by two intersecting lines.

Let line p intersect y y

the x–axis at point M. p p

Consider the angle,

, which has one side
to the right of point
M along the x–axis,
the initial side, and M x M x
the side upward
along the line, the
terminal side.

The angle, , measured in counterclockwise direction, from the initial

side to the terminal side is called the inclination of the line .
The slope of a line, denoted by is defined as the tangent of its
inclination, thus = = .

Example 1. Draw a line through P(1, 1) with slope ¾

Solution: = = , y
thus x = 4 and y = 3. By
moving 4 units to the
right of P and then 3
units upward, the
desired point was
P
obtained which will
x
then be connected by a
line to the point P. This
line drawn through
point P1 has the
required slope.

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 Example 2. Draw a line through P (-3, 3) with slope -½

Solution: = = , y
thus x = 2 and y = -1. By
moving 2 units to the right of
P and then 1 unit downward,
the desired point was
P
obtained which will then be
connected by a line to the x
point P. This line drawn
through point P has the
required slope.

If the coordinates of two points of a line are known, the slope of a line
can be found.

y
Let P1(x1, y1) and
P2(x2, y2) be two points P2(x2, y2)
on a line with slope m.
y2 - y1
Then,
= = x2 - x1
P1(x1, y1) R(x2, y1)
=
x

Again let P1(x1, y1) and y

P2(x2, y2) be two points on a
line. Now, consider the line P2(x2, y2)
which leans to the left, shown
in the figure at the right. y2 - y1

Note that and are

supplementary angles, thus x1 - x2
t n = t n . R(x2, y1) P1(x1, y1)

Since, =

then = = x

18
 Theorem: The slope of a line passing through two given points
P1(x1, y1) and P2(x2, y2) is equal to the difference of the
ordinates divided by the difference of the abscissas taken in
the same order, that is

= o =

This formula yields the slope if two points lie on a slant or a

horizontal line. If the line is vertical, the denominator will become zero,
which proves the fact that the slope of a vertical line is undefined.

Example 3: Find the slope of a line passing through the points

(1, 12) and (-2, -2).
Solution: Let P1(x1 , y1) be (1, 12)  x1 = 1 and y1 = 12
and P2(x2 , y2) be (-2, -2)  x2 = -2 and y2 = -2

= = = =

Example 4: Find the slope of a line passing through the points (0, -1)
and (4, 8).
Solution: Let P1(x1 , y1) be (0, -1)  x1 = 0 and y1 = -1
and P2(x2 , y2) be (4, 8)  x2 = 4 and y2 = 8

(
= = =

Parallel and Perpendicular Lines

Two non-vertical lines are parallel, if and only if, their slopes are
equal.
If 𝓂1 and 𝓂2 are the slopes of the lines 𝓁1 and 𝓁2 respectively,
and if 𝓂1 = 𝓂2, then 𝓁1‖𝓁2.

Two lines are perpendicular, if and only if, the slope of one is the
negative reciprocal of the slope of the other.

If 𝓂1 and 𝓂2 are the slopes of the lines 𝓁1 and 𝓁2 respectively,

and if 𝓂 = 𝓂
, then 𝓁1  𝓁2.

19
 Example 5: Show that the line 𝓁1 through the points M(2, -6) and
N(5, 2) is parallel to the line 𝓁2 through the points
O(-7, -3) and Q(-4, 5).

Solution: To show that 𝓁1‖𝓁2, should be equal to

( (
= = = =
(

Since = = , then 𝓁1‖𝓁2.

Example 6: The slope of a line perpendicular to the line through (4,

2) and (-3, y) is -. Find the value of y.

Solution: 𝓁1  𝓁2, if 𝓂 = 𝓂
Let 𝓂 be the slope of the line through (4, 2) and (-
3, y) and 𝓂 = .
Since 𝓂 = = = and 𝓂 = = ,
𝓂

then = .
Solving for y: =
-28 = 7(y – 2)
-28 = 7y – 14
7y = -14 + 28
7y = 14
y=2

Example 7: Prove by means of slope that quadrilateral with vertices

A(0, 7), B(3, 3), C(-4, 3) and D(-1, -1) is a parallelogram.

Solution: A parallelogram is a quadrilateral where opposite

sides are parallel.

= = = = =

= = = = =
(

Since = = and = = , then

AB‖CD and AC‖BD. Therefore, quadrilateral ABCD is
a parallelogram.

20
Example 8: Verify that the points X(-1, 3), Y (0, 5) and Z(3, 1) are
vertices of a right triangle.

Solution: Since a right triangle has perpendicular legs, then

slope of one leg is equal to the negative reciprocal of
the other.

= = =
(

= =

= = =
(

Since = = , then XZ  XY. Therefore, the

points X, Y and Z are vertices of a right triangle.

21
22
 EXERCISES 1.3

Name: _____________________________________________________________ Score: ______________

Section: ____________________________________________________________ Date: ____________________________

I. Find the slope of a line passing through the given pairs of points
1. (1, 9); (1, 1)

2. (1, 2); (-2, -2)

3. (0, -1); (-2, 3)

II. Show that the points A(-1, -2), B(2, 1), C(-3, 6) are vertices of a right triangle.

23
 III. Show that the points A(-2, -6) B(4, -3), C(3, -1), D(-3, -4) are vertices of a
rectangle.

24
1.4 ANGLE BETWEEN TWO LINES

This section aims to determine the measure of angles formed by

two intersecting lines.

Two interesting lines form two pairs of equal angles, where

one pair of angle is the supplement of the other pair.

In the figure at the right, y

recall that an exterior of a
𝓁2 𝓁1
triangle is equal to the sum of
the remote interior angles,
thus
=
or =
Using the formula for the
tangent of the difference of
two angles, then
x
t n = t n( – )

Let =t n nd =t n , then

where is the slope of 𝓁1, the initial side

is the slope of 𝓁2, the terminal side
is the angle measured in a counterclockwise direction
and since is the supplement of , then

= =

25
 Example: Find the tangents of the angle of △TUV with vertices at
points T(-5, 8), U(3, -2) and V(4,5).

Solution:
y

For the slope of each side of the triangle:

= = =
(

= = =
(

(
= = =

For the tangent of the angle:

( )
t n = =  =
( )( )

t n = =  =
( )

( )
t n = =  =
( )

26
 EXERCISES 1.4

Name: _____________________________________________________________ Score: ______________

Section: ____________________________________________________________ Date: ____________________________

Draw the triangle with the given vertices and find the tangents of its angle

1. U(1, -2), V(-3, 4), W(5, 6) y

27
 2. M(5, 0), N(-1, -4) and L(0, 6)
y

28
1.5. DIVISION OF A LINE SEGMENT

This section aims to determine the coordinate of a point in the

division of a line segment.

It is necessary to determine the coordinates of a point P 0 which

lies on a segment P1P2 such that the directed distance P1P0 and
P1P2 bear a certain ratio r to one another. That is

0
r= ⇒ P1P0 = r(P1P2)

y
P2(x2,y2)
Consider the segment P1P2 in
coordinate plane as shown in the P0(x0,y0)
figure at the right.
Let P0(x0, y0) be any point
P1(x1,y1) Q(x0,y1) R(x2,y1)
between P1(x1, y1) and P2(x2, y2).
x
Two similar triangles we e fo med, i.e. △P1P0Q ~ △P1P2R,
hence
0 0
= =r

where r is the ratio in which P0 divides the segment P1P2.

The statement above, in terms of the coordinates of the points,

is equivalent to
𝑥0 𝑋 𝑦0 𝑦
= 𝑟 and =𝑟
𝑥 𝑥 𝑦 𝑦

Solving for x0 and y0 from these equations,

x0 = x1 + r(x2 x1) and y0 = y1 + r(y2 y1)

29
 This leads to a theorem that. If P1(x1, y1) and P2(x2, y2) are any
two points in coordinate plane, then the coordinates of the point
0
P0(x0, y0) that divides the segment P 1P2 in the ratio r= is given
by
P0(x0, y0) = P0(x1 + r(x2 – x1), y1 + r(y2 – y1))

Although this theorem is frequently used for the case in which

P0 lies between P1 and P2. It is also valid for the case in which P0
lies on the extension (in either direction) of the segment P1P2. For
this fact the following cases were consider:

Case I. If P0 lies between P1 and P2 then 0 < r < 1.

Case II. If P0 lies on the extension of P1P2 through P2, then r > 1.
Case III. If P0 lies on the extension of P1P2 through P1, then r > 1

P2 P0 P2
P0 P1
P2
P1 P1 P0

Case I Case II Case III

Example 1: Find the coordinates of the point P0 that is three-fourths

of the way from P1 (-4, -1) to P2 (4, 3).

Solution: Since P1P0 = ¾ P1P2, then r=¾. Thus,

x0 = -4 + ¾ [4 – (-4)] = 2
and
y0 = -1 + ¾ [3 – (-1)] = 2

Therefore, the required point is (2, 2).

30
Example 2: The segment joining P1(2, -3) and P2(-8, 7) is produced
beyond P1 to a point P0 so that P0 is six times as far from
P2 as from P1 . Find the coordinates of P0.

Solution: By the statement of the problem, P0P2 = 6(P0P1). This

means that P0P1 = P1P2 or P1P0 = P1P2. Hence r = .
Thus,

x0 = 2 + ( )[(-8) – 2] = 4
and
y0 = -3 + ( )[7 – (-3)] = -5.

Therefore, the required point is (4, -5).

31
32
 EXERCISES 1.5

Name: _____________________________________________________________ Score: ______________

Section: ___________________________________________________________ Date: ____________________________

1. The points A(2, -4), B(8, 4) and C(0, 6) are the vertices of a triangle. The
line segment joining the vertex of a triangle and the midpoint of the
opposite side is called the median. Find the coordinates of the point on
each median which is of the way from the vertex to the midpoint of the
opposite side.

33
 2. Find the coordinates of the point which is 2/3 of the distance from (4, -1)
and (-1, 5) and (-1, 5).

3. Find the coordinates of P if its divides the line segment A(1, -5) B(7, -2)
so that AP:PB = 3:5

34
1.6. ANALYTIC PROOFS OF GEOMETRIC THEOREMS

This section aims to:

1. determine the area of a triangle given its three vertices; and
2. give an analytic proofs to geometric theorems.

Area of Triangle

The area of the triangle with vertices at ( , , ( , ,

( , is given by the formula

= | |

A = {[(x1y2) + (x2y3) + (x3y1)] – [(y1x2) + (y2x3) + (y3x1)]}

Note that the formula will give a positive or negative answer according to the
order of , , as it moves counterclockwise or clockwise. Thus to avoid
confusion, it is best to arrange the vertices in the counterclockwise order, so that
the formula will always yields a positive result.

Example 1: Find the area of △TUV with vertices at points

T(-5, 8), U(3, -2) and V(4, 5).

Solution: Plot the points as shown in the figure below.

35
 Moving in a counterclockwise direction, let the
points T(-5, 8), U(3, -2) and V(4, 5) be ( , ,
( , , ( , respectively. Thus,

= | |= | |

A = {[(-5)(-2)+3(5)+4(8)]–[8(3)+(-2)(-4)+5(-5)]}

= (50)

A = 25 square units

Example 2: Find the area of triangle with vertices at points

(-2, -4), (1, 6) and (5, 2).

Solution: Plot the points as shown in the figure below.

P1

P3

P2

Moving in a counterclockwise direction, let (1, 6),

(-2, -4) and (5, 2) be ( , , ( , ,
( , respectively. Thus,

36
 = | |= | |

A = {[1(-4) +(-2)2+5(6)] – [5(-4)+(-2)6+1(2)]}

= (52)

A = 26 square units

Analytic Proofs of Geometric Theorems

By the use of coordinate system, many theorems of geometry can
proved with simplicity and directress.

Example 1: Prove that the midpoint of the hypotenuse of a right

triangle is equidistant from the three vertices.

Proof: First draw a right triangle and determine the

coordinates its vertices.

P1 (0, a)

x
O (0, 0) P2 (b, 0)

First find the coordinates of the midpoint of the

hypotenuse, P1P2.

( ,  (,

37
 Let P be the midpoint of the hypotenuse, thus ( , ).
To show that ( , ) is equidistant from the vertices of the
triangle, the undirected distances of OP, PP1 and PP2 should
be equal.

OP = √( ) ( )

= √( ) ( )

=√

=√
√
=

PP1 = √( ) ( )

= √( ) ( )

=√

=√
√
=

PP2 = √( ) ( )

= √( ) ( )

=√

=√
√
=

Since the undirected distances of OP, PP1 and PP2 are all
√
equal to , then the theorem is proved.

38
Example 2: Prove that the diagonals of a parallelogram bisect each
other.
Proof: First draw a parallelogram and determine the
coordinates of its vertices.

P1 (a, c) P2 (a+b, c)

O (0, 0) P3 (b, 0)

Note that a segment bisector passes through the

midpoint of the bisected segment, thus to show that the
diagonals, nd , bisect each other, the coordinates of
the midpoint of each diagonal should be found.
For the midpoint of OP2: For the midpoint of P1P3:
(
( , ( ,

( , ( ,

Since the midpoint of diagonals had the same coordinates,

( ) and ( , then the diagonals bisect
each other. Therefore, the theorem is proved.

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 EXERCISES 1.6

Name: _____________________________________________________________ Score: ______________

Section: ___________________________________________________________ Date: ____________________________

I. Find the area of the triangle with given vertices

1. A(-2, -3), B(3, 2) and C(1, 8)

2. D(-1, -2), E(2, 1), F(-3, 6)

3. J(-2, -6), K(4, -3), L(3, -1)

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 II. Give analytic proofs for each of the following:

1. The sum of the square of the distances of any point from two opposite
vertices of a rectangle is equal to the sum of the squares of its
distances from the other two vertices.

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2. The sum of the squares of the four sides of a parallelogram is equal to
the sum of the squares of the diagonals.

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 3. The sum of the squares of the medians of the triangle is equal to
three-fourths the sum of the squares of the sides.

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