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Geometry: Triangle geometry (Grade 11)

Module by: Free High School Science Texts Project. E-mail the author

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 A B B C = x y = k x k y = D E E F
(1)
the line segments are in the same proportion the line segments are in the same proportion
(2)

Figure 1
Figure 1 (MG11C16_009.png)

If the line segments are proportional, the following also hold

  1. C B A C = F E D F C B A C = F E D F
  2. A C · F E = C B · D F A C · F E = C B · D F
  3. ABBC=DEFEABBC=DEFE and BCAB=FEDEBCAB=FEDE
  4. ABAC=DEDFABAC=DEDF and ACAB=DFDEACAB=DFDE

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 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
(3)

Figure 2
Figure 2 (MG11C16_010.png)

  • 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.
    areaABC=12·h·BC=areaDBCareaABC=12·h·BC=areaDBC
    (4)
    Figure 3
    Figure 3 (MG11C16_011.png)
  • Triangles on the same side of the same base, with equal areas, lie between parallel lines.
    IfareaABC=areaBDC,IfareaABC=areaBDC,
    (5)
    thenADBC.thenADBC.
    (6)
    Figure 4
    Figure 4 (MG11C16_012.png)

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

Figure 5
Figure 5 (MG11C16_013.png)

Given:ABC with line DE BC

R.T.P.:

A D D B = A E E C A D D B = A E E C
(7)

Proof: Draw h1h1 from E perpendicular to AD, and h2h2 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. 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.
(8)

Similarly,

A D A B = A E A C A B B D = A C C E A D A B = A E A C A B B D = A C C E
(9)

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 = 2AB AE = AC + CB = 2AC then DE BC and BC = 2DE.

Figure 6
Figure 6 (MG11C16_014.png)

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

Figure 7
Figure 7 (MG11C16_015.png)

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

R.T.P.:

A B D E = A C D F A B D E = A C D F
(10)

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 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
(11)
Tip:
|||||| means “is similar to"

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

Figure 8
Figure 8 (MG11C16_016.png)

Given:ABC with line DE such that

A D D B = A E E C A D D B = A E E C
(12)

R.T.P.:DEBCDEBC; ADE ||||||ABC

Proof: Draw h1h1 from E perpendicular to AD, and h2h2 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 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
(13)
ADE and ABC are equiangular ADE and ABC are equiangular
(14)
A D E | | | A B C (AAA) A D E | | | A B C (AAA)
(15)

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^=90A^=90

Figure 9
Figure 9 (MG11C16_017.png)

Required to prove:BC2=AB2+AC2BC2=AB2+AC2

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 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
(16)
ABD | | | CBAand CAD | | | CBA ( AAA ) ABD | | | CBAand CAD | | | CBA ( AAA )
(17)
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 C B = B D B A = A D C A and C A C B = C D C A = A D B A
(18)
A B 2 = C B × B D and A C 2 = C B × C D A B 2 = C B × B D and A C 2 = C B × C D
(19)
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 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
(20)
Exercise 1: Triangle Geometry 1

In GHI, GH LJ; GJ LK and JKKIJKKI = 5353. Determine HJKIHJKI.

Figure 10
Figure 10 (MG11C16_018.png)

Solution
  1. Step 1. Identify similar triangles :
    L I ^ J = G I ^ H J L ^ I = H G ^ I ( Corres . s ) L I J | | | G I H ( Equiangular s ) L I ^ J = G I ^ H J L ^ I = H G ^ I ( Corres . s ) L I J | | | G I H ( Equiangular s )
    (21)
    L I ^ K = G I ^ J K L ^ I = J G ^ I ( Corres . s ) L I K | | | G I J ( Equiangular s ) L I ^ K = G I ^ J K L ^ I = J G ^ I ( Corres . s ) L I K | | | G I J ( Equiangular s )
    (22)
  2. Step 2. Use proportional sides :
    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 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
    (23)
  3. Step 3. Rearrange to find the required ratio :
    H J K I = H J J I × J I K I H J K I = H J J I × J I K I
    (24)

    We need to calculate JIKIJIKI: We were given JKKI=53JKKI=53 So rearranging, we have JK=53KIJK=53KI And:

    J I = J K + K I = 5 3 K I + K I = 8 3 K I J I K I = 8 3 J I = J K + K I = 5 3 K I + K I = 8 3 K I J I K I = 8 3
    (25)

    Using this relation:

    = 5 3 × 8 3 = 40 9 = 5 3 × 8 3 = 40 9
    (26)
Exercise 2: Triangle Geometry 2

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

Figure 11
Figure 11 (MG11C16_019.png)

Solution
  1. Step 1. Identify similar triangles :
    P 1 ^ = S 1 ^ ( Alt . s ) Q 1 ^ = R 1 ^ ( Alt . s ) P T Q | | | S T R ( Equiangular s ) P 1 ^ = S 1 ^ ( Alt . s ) Q 1 ^ = R 1 ^ ( Alt . s ) P T Q | | | S T R ( Equiangular s )
    (27)
  2. Step 2. Use proportional sides :
    P T T Q = S T T R ( P T Q | | | S T R ) P T · T R = S T · T Q P T T Q = S T T R ( P T Q | | | S T R ) P T · T R = S T · T Q
    (28)
Triangle Geometry
  1. Calculate SV
    Figure 12
    Figure 12 (MG11C16_020.png)
  2. CBYB=32CBYB=32. Find DSSBDSSB.
    Figure 13
    Figure 13 (MG11C16_021.png)
  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.
    Figure 14
    Figure 14 (MG11C16_022.png)
  4. Using the following figure and lengths, find IJ and KJ. HI = 26 m, KL = 13 m, JL = 9 m and HJ = 32 m.
    Figure 15
    Figure 15 (MG11C16_023.png)
  5. Find FH in the following figure.
    Figure 16
    Figure 16 (MG11C16_024.png)
  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 DEACDEAC.
    Figure 17
    Figure 17 (MG11C16_025.png)
  7. If LM JK, calculate yy.
    Figure 18
    Figure 18 (MG11C16_026.png)

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