4.1 Triangle geometry

Triangle geometry

Proportion

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

If the line segments are proportional, the following also hold

1. $\frac{CB}{AC}=\frac{FE}{DF}$
2. $AC·FE=CB·DF$
3. $\frac{AB}{BC}=\frac{DE}{FE}$ and $\frac{BC}{AB}=\frac{FE}{DE}$
4. $\frac{AB}{AC}=\frac{DE}{DF}$ and $\frac{AC}{AB}=\frac{DF}{DE}$

Proportionality of triangles

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

• 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.
  $$\text{area}▵ABC=\frac{1}{2}·h·BC=\text{area}▵DBC$$

  

• Triangles on the same side of the same base, with equal areas, lie between parallel lines.
  $$\text{If}\text{area}▵\text{ABC}\text{=}\text{area}▵\text{BDC,}$$
  $$\text{then}\text{AD}\parallel \text{BC.}$$

  

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 $\parallel$ BC

R.T.P. :

Proof : Draw $h_1$ from E perpendicular to AD, and $h_2$ from D perpendicular to AE.

Draw BE and CD.

Similarly,

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 $\parallel$ 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 $\widehat{A}=\widehat{D}$ ; $\widehat{B}=\widehat{E}$ ; $\widehat{C}=\widehat{F}$

R.T.P. :

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

Proof : In $▵$ 's AGH and DEF

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

Given : $▵$ ABC with line DE such that

R.T.P. : $DE\parallel BC$ ; $▵$ ADE $\left|\right||$ $▵$ ABC

Proof : Draw $h_1$ from E perpendicular to AD, and $h_2$ from D perpendicular to AE.

Draw BE and CD.

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 $\widehat{A}={90}^{\circ }$

Required to prove : $B{C}^2=A{B}^2+A{C}^2$

Proof :

In $▵$ GHI, GH $\parallel$ LJ; GJ $\parallel$ LK and $\frac{JK}{KI}$ = $\frac{5}{3}$ . Determine $\frac{HJ}{KI}$ .

1. Identify similar triangles
   $$\begin{array}{ccc}\hfill L\widehat{I}J& =& G\widehat{I}H\hfill \\ \hfill J\widehat{L}I& =& H\widehat{G}I(\mathrm{Corres}.\angle \mathrm{s})\hfill \\ \hfill \therefore ▵LIJ& \left|\right||& ▵GIH(\mathrm{Equiangular}▵\mathrm{s})\hfill \end{array}$$
   $$\begin{array}{ccc}\hfill L\widehat{I}K& =& G\widehat{I}J\hfill \\ \hfill K\widehat{L}I& =& J\widehat{G}I(\mathrm{Corres}.\angle \mathrm{s})\hfill \\ \hfill \therefore ▵LIK& \left|\right||& ▵GIJ(\mathrm{Equiangular}▵\mathrm{s})\hfill \end{array}$$
2. Use proportional sides
   $$\begin{array}{ccc}\hfill \frac{HJ}{JI}& =& \frac{GL}{LI}\left(▵LIJ\right|\left|\right|▵GIH)\hfill \\ \hfill \text{and}\frac{GL}{LI}& =& \frac{JK}{KI}\left(▵LIK\right|\left|\right|▵GIJ)\hfill \\ & =& \frac{5}{3}\hfill \\ \hfill \therefore \frac{HJ}{JI}& =& \frac{5}{3}\hfill \end{array}$$
3. Rearrange to find the required ratio
   $$\begin{array}{ccc}\hfill \frac{HJ}{KI}& =& \frac{HJ}{JI}\times \frac{JI}{KI}\hfill \end{array}$$

   We need to calculate $\frac{JI}{KI}$ : We were given $\frac{JK}{KI}=\frac{5}{3}$ So rearranging, we have $JK=\frac{5}{3}KI$ And:

   $$\begin{array}{ccc}\hfill JI& =& JK+KI\hfill \\ & =& \frac{5}{3}KI+KI\hfill \\ & =& \frac{8}{3}KI\hfill \\ \hfill \frac{JI}{KI}& =& \frac{8}{3}\hfill \end{array}$$

   Using this relation:

   $$\begin{array}{ccc}& =& \frac{5}{3}\times \frac{8}{3}\hfill \\ & =& \frac{40}{9}\hfill \end{array}$$

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

1. Identify similar triangles
   $$\begin{array}{ccc}\hfill \widehat{P_1}& =& \widehat{S_1}(\mathrm{Alt}.\angle \mathrm{s})\hfill \\ \hfill \widehat{Q_1}& =& \widehat{R_1}(\mathrm{Alt}.\angle \mathrm{s})\hfill \\ \hfill \therefore ▵PTQ& \left|\right||& ▵STR(\mathrm{Equiangular}▵\mathrm{s})\hfill \end{array}$$
2. Use proportional sides
   $$\begin{array}{ccc}\hfill \frac{PT}{TQ}& =& \frac{ST}{TR}\left(▵PTQ\right|\left|\right|▵STR)\hfill \\ \hfill \therefore PT·TR& =& ST·TQ\hfill \end{array}$$

Triangle geometry

1. Calculate SV

   

2. $\frac{CB}{YB}=\frac{3}{2}$ . Find $\frac{DS}{SB}$ .

   

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 $\frac{DE}{AC}$ .

   

7. If LM $\parallel$ JK, calculate $y$ .