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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 .

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Source:  OpenStax, Siyavula textbooks: grade 11 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11243/1.3
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