<< Chapter < Page Chapter >> Page >

Triangle geometry

Proportion

Two line segments are divided in the same proportion if the ratios between their parts are equal.

A B B C = x y = k x k y = D E E F
the line segments are in the same proportion

If the line segments are proportional, the following also hold

  1. C B A C = F E D F
  2. A C · F E = C B · D F
  3. A B B C = D E F E and B C A B = F E D E
  4. A B A C = D E D F and A C A B = D F D E

Proportionality of triangles

Triangles with equal heights have areas which are in the same proportion to each other as the bases of the triangles.

h 1 = h 2 area A B C area D E F = 1 2 B C × h 1 1 2 E F × h 2 = B C E F

  • A special case of this happens when the bases of the triangles are equal: Triangles with equal bases between the same parallel lines have the same area.
    area A B C = 1 2 · h · B C = area D B C
  • Triangles on the same side of the same base, with equal areas, lie between parallel lines.
    If area ABC = area BDC,
    then AD BC.

Theorem 1 Proportion Theorem: A line drawn parallel to one side of a triangle divides the other two sides proportionally.

Given : ABC with line DE BC

R.T.P. :

A D D B = A E E C

Proof : Draw h 1 from E perpendicular to AD, and h 2 from D perpendicular to AE.

Draw BE and CD.

area ADE area BDE = 1 2 A D · h 1 1 2 D B · h 1 = A D D B area ADE area CED = 1 2 A E · h 2 1 2 E C · h 2 = A E E C but area BDE = area CED (equal base and height) area ADE area BDE = area ADE area CED A D D B = A E E C DE divides AB and AC proportionally.

Similarly,

A D A B = A E A C A B B D = A C C E

Following from Theorem  "Proportion" , we can prove the midpoint theorem.

Theorem 2 Midpoint Theorem: A line joining the midpoints of two sides of a triangle is parallel to the third side and equal to half the length of the third side.

Proof : This is a special case of the Proportionality Theorem (Theorem "Proportion" ). If AB = BD and AC = AE,and AD = AB + BD = 2ABAE = AC + CB = 2AC then DE BC and BC = 2DE.

Theorem 3 Similarity Theorem 1: Equiangular triangles have their sides in proportion and are therefore similar.

Given : ABC and DEF with A ^ = D ^ ; B ^ = E ^ ; C ^ = F ^

R.T.P. :

A B D E = A C D F

Construct: G on AB, so that AG = DE, H on AC, so that AH = DF

Proof : In 's AGH and DEF

AG = DE (const.) AH = D ( const. ) A ^ = D ^ ( given ) AGH DEF ( SAS ) A G ^ H = E ^ = B ^ G H BC ( corres. 's equal ) AG AB = A H A C ( proportion theorem ) DE AB = D F A C ( AG = DE ; AH = DF ) ABC | | | DEF
| | | means “is similar to"

Theorem 4 Similarity Theorem 2: Triangles with sides in proportion are equiangular and therefore similar.

Given : ABC with line DE such that

A D D B = A E E C

R.T.P. : D E B C ; ADE | | | ABC

Proof : Draw h 1 from E perpendicular to AD, and h 2 from D perpendicular to AE.

Draw BE and CD.

area ADE area BDE = 1 2 A D · h 1 1 2 D B · h 1 = A D D B area ADE area CED = 1 2 A E · h 2 1 2 E C · h 2 = A E E C but A D D B = A E E C (given) area ADE area BDE = area ADE area CED area BDE = area CED D E B C (same side of equal base DE, same area) A D ^ E = A B ^ C (corres 's) and A E ^ D = A C ^ B
ADE and ABC are equiangular
A D E | | | A B C (AAA)

Theorem 5 Pythagoras' Theorem: The square on the hypotenuse of a right angled triangle is equal to the sum of the squares on the other two sides.

Given : ABC with A ^ = 90

Required to prove : B C 2 = A B 2 + A C 2

Proof :

Let C ^ = x D A ^ C = 90 - x ( 's of a ) D A ^ B = x A B ^ D = 90 - x ( 's of a ) B D ^ A = C D ^ A = A ^ = 90
ABD | | | CBA and CAD | | | CBA ( AAA )
A B C B = B D B A = A D C A and C A C B = C D C A = A D B A
A B 2 = C B × B D and A C 2 = C B × C D
A B 2 + A C 2 = C B ( B D + C D ) = C B ( C B ) = C B 2 i . e . B C 2 = A B 2 + A C 2

In GHI, GH LJ; GJ LK and J K K I = 5 3 . Determine H J K I .

  1. L I ^ J = G I ^ H J L ^ I = H G ^ I ( Corres . s ) L I J | | | G I H ( Equiangular s )
    L I ^ K = G I ^ J K L ^ I = J G ^ I ( Corres . s ) L I K | | | G I J ( Equiangular s )
  2. H J J I = G L L I ( L I J | | | G I H ) and G L L I = J K K I ( L I K | | | G I J ) = 5 3 H J J I = 5 3
  3. H J K I = H J J I × J I K I

    We need to calculate J I K I : We were given J K K I = 5 3 So rearranging, we have J K = 5 3 K I And:

    J I = J K + K I = 5 3 K I + K I = 8 3 K I J I K I = 8 3

    Using this relation:

    = 5 3 × 8 3 = 40 9

PQRS is a trapezium, with PQ RS. Prove that PT · TR = ST · TQ.

  1. P 1 ^ = S 1 ^ ( Alt . s ) Q 1 ^ = R 1 ^ ( Alt . s ) P T Q | | | S T R ( Equiangular s )
  2. P T T Q = S T T R ( P T Q | | | S T R ) P T · T R = S T · T Q

Triangle geometry

  1. Calculate SV
  2. C B Y B = 3 2 . Find D S S B .
  3. Given the following figure with the following lengths, find AE, EC and BE. BC = 15 cm, AB = 4 cm, CD = 18 cm, and ED = 9 cm.
  4. Using the following figure and lengths, find IJ and KJ. HI = 26 m, KL = 13 m, JL = 9 m and HJ = 32 m.
  5. Find FH in the following figure.
  6. BF = 25 m, AB = 13 m, AD = 9 m, DF = 18m. Calculate the lengths of BC, CF, CD, CE and EF, and find the ratio D E A C .
  7. If LM JK, calculate y .

Questions & Answers

general equation for photosynthesis
Ojasope Reply
6CO2 + 6H2O + solar energy= C6H1206+ 6O2
Anastasiya
meaning of amino Acids
AJAYI Reply
a diagram of an adult mosquito
mubarak Reply
what are white blood cells
Mlungisi Reply
white blood cell is part of the immune system. that help fight the infection.
MG
what about tissue celss
Mlungisi
Cells with a similar function, form a tissue. For example the nervous tissue is composed by cells:neurons and glia cells. Muscle tissue, is composed by different cells.
Anastasiya
I need further explanation coz celewi anything guys,,,
Calvin Reply
hey guys
Isala
on what?
Anastasiya
hie
Lish
is air homogenous or hetrogenous
damiane Reply
homogenous
Kevin
why saying homogenous?
Isala
explain if oxygen is necessary for photosynthesis
Allice Reply
explain if oxygen is necessary for photosynthesis
Allice Reply
Yes, the plant does need oxygen. The plant uses oxygen, water, light, and produced food. The plant use process called photosynthesis.
MG
By using the energy of sunlight, plants convert carbon dioxide and water into carbohydrates and oxygen by photosynthesis. This happens during the day and sunlight is needed.
NOBLE
no. it s a product of the process
Anastasiya
yet still is it needed?
NOBLE
no. The reaction is: 6CO2+6H20+ solar energy =C6H12O6(glucose)+602. The plant requires Carbon dioxyde, light, and water Only, and produces glucose and oxygen( which is a waste).
Anastasiya
what was the question
NOBLE Reply
joining
Godfrey
the specific one
NOBLE
the study of non and living organism is called.
Godfrey
Is call biology
Alohan
yeah
NOBLE
yes
Usher
what Is ecology
Musonda Reply
what is a cell
Emmanuel Reply
A cell is a basic structure and functional unit of life
Ndongya
what is biolgy
Hawwi Reply
is the study of living and non living organisms
Ahmed
may u draw the female organ
MARTIN Reply
i dont understand
Asal
:/
Asal
me too
DAVID
anabolism and catabolism
Sani Reply
Anabolism refers to the process in methabolism in which complex molecules are formed "built" and requires energy to happen. Catabolism is the opposite process: complex molecules are deconstructed releasing energy, such as during glicolysis.
Anastasiya
Explain briefly independent assortment gene .
Otu Reply
hi
Amargo
hi I'm Anatalia
Joy
what do you mean by pituitary gland
Digambar
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, Siyavula textbooks: grade 11 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11243/1.3
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Siyavula textbooks: grade 11 maths' conversation and receive update notifications?

Ask