some Calculations in plane Geometry

This article describes some formula's in plane geometry.
They are nice applications of algebra.
List of contents:

• the area of a triangle
• the projection formula
• Stewart's formula
• the length of the median
• the bisector formula
• the radius of the circumscribing circle
• the radius of the inscribed circle
• the radius of the escribed circle
• kissing circles
• intersecting circles

the area of a triangle
In cases where the length of the sides of a triangle are known,
but not the height, a formula for the area of the triangle without
this height would be of help.

Please look at the picture right:
the base, opposite angle A, has length a.
BD = x, so DC = a - x. AD = h.
Side b is opposite angle B, side c is opposite angle C.

There we go:
application of the theorem of Pythagoras in triangles ABD and ADC
results in two equations:

­	­	c 2 = h 2 + x 2 b 2 = h 2 + (a − x) 2
­

or:

­	­	h 2 = c 2 − x 2 h 2 = b 2 − (a − x) 2
­

so

c ² − x ² = b ² − (a − x) ²

and

c ² − x ² = b ² − (a ² − 2 a x + x ²)

c ² − x ² = b ² − a ² + 2 a x − x ²

c ² = b ² − a ² + 2 a x

2 a x = a ² − b ² + c ²

x = 

a 2 − b 2 + c 2

2 a

this value of x, substituted in h² = c² - x² = (c - x)(c + x):

h ² = 

æ	c −  a 2 − b 2 + c 2 2 a	ö
­		­
è		ø

 

æ	c +  a 2 − b 2 + c 2 2 a	ö
­		­
è		ø

h ² = 

2 a c − a 2 + b 2 − c 2

2 a

 · 

2 a c + a 2 − b 2 + c 2

2 a

h ² = 

−(a 2 − 2 a c + c 2 − b 2)

2 a

 · 

a 2 + 2 a c + c 2 − b 2

2 a

h ² = 

−((a − c) 2 − b 2)

2 a

 · 

(a + c) 2 − b 2

2 a

h ² = 

b 2 − (a − c) 2

2 a

 · 

(a + c) 2 − b 2

2 a

h ² = 

(b − a + c) (b + a − c)

2 a

 · 

(a + c − b) (a + c + b)

2 a

h ² = 

(−a + b + c) (a + b − c) (a − b + c) (a + b + c)

4 a 2

This formula can be simplified with a tric.
If s is half of the circumference, than

( a + b + c) = 2s
(-a + b + c) = 2s - 2a
( a + b - c) = 2s - 2c
( a - b + c) = 2s - 2b

and the formula changes to

h ² = 

2 s (2 s − 2 a) (2 s − 2 b) (2 s − 2 c)

4 a 2

h ² = 

4

a 2

 s (s − a) (s − b) (s − c)

h = 

2

a

 

\	s (s − a) (s − b) (s − c)

and because tghe area A = 

a h

2

A = 

\	s (s − a) (s − b) (s − c)

with this formula, the area A of a triangle can be calculated if the sides a,b,c
are known (but not the height).

An Example
A triangle has sides of length 7, 8 and 11.
So, s = (7 + 8 + 11) / 2 = 13.
The area O = 

\	13 (13 − 7) (13 − 8) (13 − 11)

 = 27 , 928 m ²

the projection formula
Application of the Pythagoras theorem:

a² = h² + (c - p)²
a² = b² - p² + c² - 2cp + p²
a² = b² + c² - 2cp

the projection formula:

a2 = b2 + c2 - 2cp

remark: If point D is right of B, -2cp changes into +2cp.
The reader may verify this.
p is the projection of b to c. hence the name.

Stewart's formula
P is a random point on base AB.
P divides AB in lines c₁ and c₂.

The formula of Stewart calculates the length of CP, if the sides a,b,c
of the triangle and the position of P are kwown.

Application of the projection formula in triangles APC and ABC :

­	­	C P 2 = b 2 + c1 2 − 2 c1 p a 2 = b 2 + c 2 − 2 c p
­

multiplication of the first line with (c) and the second with(c₁):

­	­	c C P 2 = b 2 c + c c1 2 − 2 c c1 p c1 a 2 = b 2 c1 + c1 c 2 − 2 c c1 p
­

Subtract the lower equation from the upper:

c C P ² − c₁ a ² = b ² c − b ² c₁ + c c₁ ² − c₁ c ²

c C P ² = c₁ a ² + b ² c − b ² c₁ + c c₁ ² − c₁ c ²

c C P ² = c₁ a ² + b ² (c₁ + c₂) − b ² c₁ + c c₁ ² − c₁ c ²

c C P ² = c₁ a ² + c₂ b ² + c c₁ (c₁ − c)

c C P ² = c₁ a ² + c₂ b ² + c c₁ (c₁ − c₁ − c₂)

c C P ² = c₁ a ² + c₂ b ² − c c₁ c₂

this is Stewart's formula:

c.CP2 = c1a2 + c2b2 - 2c.c1.c2

the length of the median
If AP = PB then CP is the median of angle C.
For c₁ = c₂ and CP = Z_c Stewart's formula changes into:

c Z_c ² = 

c

2

 a ² + 

c

2

 b ² − c · 

c

2

 · 

c

2

Z_c ² = 

a 2

2

 + 

b 2

2

 − 

c 2

4

this is the formula for the median of a triangle:

Zc 2 =  a 2 2  +  b 2 2  −  c 2 4

the bisector formula
CD divides angle C in equal parts.
First a theorem of plane geometry to get a relation
between a,b and c₁,c₂.

Given:

DABC
CD divides LC.

we prove that:

In a triangle, the bisector divides the opposite side in parts with
the same ratio as the sides of the angle:
AD : DB = AC : BC

Proof:

make BC longer
draw AF || CD and
EC || AB
because of parallellism of lines:
LC₁ = LA₁
LC₂ = LF
so

LF = LA₁
in a triangle, sides opposite equal angles are also equal, so
AC = FC
because of similarity of DECF and DDBC

EC : DB = FC : BC
because ADCE is a rhombus:

EC = AD
so:

AD : DB = FC : BC
and because FC = AC
AD : DB = AC : BC

back to the bisector formula.

Starting with

a

b

 = 

c2

c1

we write:

ac₁ = bc₂

Stewart's formula

c C P ² = c₁ a ² + c₂ b ² − c c₁ c₂

is changed into

c C P ² = a b c₂ + a b c₁ − c c₁ c₂

and

c C P ² = a b (c₁ + c₂) − c c₁ c₂

C P ² = a b − c₁ c₂

The bisector formula is

C P 2 = a b − c1 c2

If c₁ and c₂ are unknown than from:

­	­	a c1 = b c2 c1 = c − c2
­

we may calculate

c₁ = 

b c

a + b


c₂ = 

a c

a + b

c₁ en c₂ substitution into the bisector formula results in

C P ² = 

4 a b

(a + b) 2

 s (s − c)

s is half of the circumference of the triangle.
Calculations are left to the reader.

the radius of a circumscribing circle
First another theorem:

If two triangles have a common angle, then the ratio of their areas is equal to
the ratio of the products of their sides forming the angle.

DABC and DADE have LA in common.

area.DABC : area.DADC = (p.AB) : (p.AD) = AB : AD
area.DADC : area.DADE = (q.AC) : (q.AE) = AC : AE

multiplication by AC and AD:

AB : AD = (AB.AC) : (AD.AC)
AC : AE = (AD.AC) : (AD.AE)

so

area.DABC : area.DADE = (AB.AC) : (AD.AE)

In the picture right, M is the middle of
the circumscribing circle of DABC.

applying the theorem of Thales:

LDMB = 0.5*LAMB = LC

so:

area DABC : area DDMB = (AC.BC) : (DM.BM)

but also:

area DABC : area DDMB = (

c

2

 h ) : (

c

4

 D M )

because

BM = R {the radius}
AC = b
BC = a
AB = c

follows:

a b D M.R

=

c 2  h c 4  D M

and

R = 

a b

2 h

we know that area DABC = 0.5hc, so

h = 2*area DABC / c and

the length of radius R of the circle is:

R =  a b c 4 A

where A is the area of the triangle ABC.

the radius of inscribed circle

2 * area DABC = ar + br + cr = r(a + b + c)

so:

r =  A s

Remark: s is half the circumference of the triangle.

the radius of the escribed circle
The radius r_a is

ra =  A s − a

where:

BC = a
AC = b
AB = c
(a + b + c) / 2 = s

The proof is left to the reader.

Suggested exercises:

r, r_a , r_b , r_c are the radius of the in- and escribes circles of DABC.
A is the area of DABC.
prove that:

1. A = 

\	r ra rb rc

2.

1

r

 = 

1

ra

 + 

1

rb

 + 

1

rc

kissing circles
Two circles with radius r and R are placed on a line.
We derive a formula for the distance d = AB.

Application of the theorem of Pythagoras:

(R + r)² = d² + (R - r)²
R² + 2Rr + r² = d² + R² - 2Rr + r²

d = 2  \ R r

Suggested exercise:

In the picture above, a small circle with radius x just fits between the
line and the circles with radius r and R.

3. using the derived formula for d, prove that

1 \ x

=

1 \ R  +  1 \ r

4. see picture right.
P is a fixed point.
A and B are randomly choosen points on the circle.
Prove, that PA.PB is a constant value.
(called: the power of P relative to the circle)

intersecting circles
Circles with radius R and r and d = MN < R + r intersect.
See picture right.
We derive a formula for the length of MP.

The picture shows

MT = d - r
NU = d - R

We add a point Q on line MN such, that MQ = QN.

Now

R² - MP² = r² - NP² or
MP² - NP² = R² - r²
(MP + NP)(MP - NP) = (d)(MQ + QP - (QN - QP)) = 2d.QP

because

2d.QP = R² - r²
Q P = 

R 2 − r 2

2 d


M P = M Q + Q P = 

d

2

 + Q P

M P =  d 2 + (R 2 − r 2) 2 d

Suggested exercise:
see picture above.
5. find also formula's for NP, TP, UP.

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