# 4.1 Triangle geometry

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## Proportion

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

$\frac{AB}{BC}=\frac{x}{y}=\frac{kx}{ky}=\frac{DE}{EF}$
$\therefore \text{the}\phantom{\rule{4.pt}{0ex}}\text{line}\phantom{\rule{4.pt}{0ex}}\text{segments}\phantom{\rule{4.pt}{0ex}}\text{are}\phantom{\rule{4.pt}{0ex}}\text{in}\phantom{\rule{4.pt}{0ex}}\text{the}\phantom{\rule{4.pt}{0ex}}\text{same}\phantom{\rule{4.pt}{0ex}}\text{proportion}$

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.

$\begin{array}{ccc}\hfill {h}_{1}& =& {h}_{2}\hfill \\ \hfill \therefore \frac{\text{area}\phantom{\rule{4.pt}{0ex}}▵ABC}{\text{area}\phantom{\rule{4.pt}{0ex}}▵DEF}& =& \frac{\frac{1}{2}BC×{h}_{1}}{\frac{1}{2}EF×{h}_{2}}=\frac{BC}{EF}\hfill \end{array}$

• 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.
$\text{area}\phantom{\rule{4.pt}{0ex}}▵ABC=\frac{1}{2}·h·BC=\phantom{\rule{4.pt}{0ex}}\text{area}\phantom{\rule{4.pt}{0ex}}▵DBC$
• Triangles on the same side of the same base, with equal areas, lie between parallel lines.
$\text{If}\phantom{\rule{4.pt}{0ex}}\text{area}\phantom{\rule{4.pt}{0ex}}▵\phantom{\rule{4.pt}{0ex}}\text{ABC}\phantom{\rule{4.pt}{0ex}}\text{=}\phantom{\rule{4.pt}{0ex}}\text{area}\phantom{\rule{4.pt}{0ex}}▵\phantom{\rule{4.pt}{0ex}}\text{BDC,}$
$\text{then}\phantom{\rule{4.pt}{0ex}}\text{AD}\phantom{\rule{4.pt}{0ex}}\parallel \phantom{\rule{4.pt}{0ex}}\text{BC.}$

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. :

$\frac{AD}{DB}=\frac{AE}{EC}$

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

Draw BE and CD.

$\begin{array}{ccc}\hfill \frac{\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{ADE}}{\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{BDE}}& =& \frac{\frac{1}{2}AD·{h}_{1}}{\frac{1}{2}DB·{h}_{1}}=\frac{AD}{DB}\hfill \\ \hfill \frac{\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{ADE}}{\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{CED}}& =& \frac{\frac{1}{2}AE·{h}_{2}}{\frac{1}{2}EC·{h}_{2}}=\frac{AE}{EC}\hfill \\ \hfill \text{but}\phantom{\rule{4.pt}{0ex}}\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{BDE}\phantom{\rule{4.pt}{0ex}}& =& \phantom{\rule{4.pt}{0ex}}\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{CED}\phantom{\rule{4.pt}{0ex}}\text{(equal}\phantom{\rule{4.pt}{0ex}}\text{base}\phantom{\rule{4.pt}{0ex}}\text{and}\phantom{\rule{4.pt}{0ex}}\text{height)}\hfill \\ \hfill \therefore \frac{\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{ADE}}{\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{BDE}}& =& \frac{\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{ADE}}{\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{CED}}\hfill \\ \hfill \therefore \frac{AD}{DB}& =& \frac{AE}{EC}\hfill \\ \hfill \therefore \text{DE}\phantom{\rule{4.pt}{0ex}}\text{divides}\phantom{\rule{4.pt}{0ex}}\text{AB}\phantom{\rule{4.pt}{0ex}}\text{and}\phantom{\rule{4.pt}{0ex}}\text{AC}\phantom{\rule{4.pt}{0ex}}\text{proportionally.}\end{array}$

Similarly,

$\begin{array}{ccc}\hfill \frac{AD}{AB}& =& \frac{AE}{AC}\hfill \\ \hfill \frac{AB}{BD}& =& \frac{AC}{CE}\hfill \end{array}$

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 $\stackrel{^}{A}=\stackrel{^}{D}$ ; $\stackrel{^}{B}=\stackrel{^}{E}$ ; $\stackrel{^}{C}=\stackrel{^}{F}$

R.T.P. :

$\frac{AB}{DE}=\frac{AC}{DF}$

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

Proof : In $▵$ 's AGH and DEF

$\begin{array}{cccc}\hfill \mathrm{AG}& =& \mathrm{DE}\hfill & \mathrm{\left(const.\right)}\\ \hfill \mathrm{AH}& =& D\hfill & \left(\mathrm{const.}\right)\\ \hfill \stackrel{^}{A}& =& \stackrel{^}{D}\hfill & \left(\mathrm{given}\right)\\ \hfill \therefore \phantom{\rule{3pt}{0ex}}▵\mathrm{AGH}& \equiv & ▵\mathrm{DEF}\hfill & \left(\mathrm{SAS}\right)\\ \hfill \therefore \phantom{\rule{3pt}{0ex}}A\stackrel{^}{G}H& =& \stackrel{^}{E}=\stackrel{^}{B}\hfill & \\ \therefore \phantom{\rule{3pt}{0ex}}GH\hfill & \parallel & \mathrm{BC}\hfill & \left(\mathrm{corres.}\angle \mathrm{\text{'}s equal}\right)\\ \hfill \therefore \phantom{\rule{3pt}{0ex}}\frac{\mathrm{AG}}{\mathrm{AB}}& =& \frac{AH}{AC}\hfill & \left(\mathrm{proportion theorem}\right)\\ \hfill \therefore \phantom{\rule{3pt}{0ex}}\frac{\mathrm{DE}}{\mathrm{AB}}& =& \frac{DF}{AC}\hfill & \left(\mathrm{AG}=\mathrm{DE};\mathrm{AH}=\mathrm{DF}\right)\\ \hfill \therefore \phantom{\rule{3pt}{0ex}}▵\mathrm{ABC}& |||& ▵\mathrm{DEF}\hfill \end{array}$
$|||$ 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

$\frac{AD}{DB}=\frac{AE}{EC}$

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

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

Draw BE and CD.

$\begin{array}{ccc}\hfill \frac{\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{ADE}}{\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{BDE}}& =& \frac{\frac{1}{2}AD·{h}_{1}}{\frac{1}{2}DB·{h}_{1}}=\frac{AD}{DB}\hfill \\ \hfill \frac{\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{ADE}}{\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{CED}}& =& \frac{\frac{1}{2}AE·{h}_{2}}{\frac{1}{2}EC·{h}_{2}}=\frac{AE}{EC}\hfill \\ \hfill \text{but}\phantom{\rule{4.pt}{0ex}}\frac{AD}{DB}& =& \frac{AE}{EC}\phantom{\rule{4.pt}{0ex}}\text{(given)}\hfill \\ \hfill \therefore \frac{\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{ADE}}{\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{BDE}}& =& \frac{\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{ADE}}{\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{CED}}\hfill \\ \hfill \therefore \text{area}\phantom{\rule{4.pt}{0ex}}▵\text{BDE}\phantom{\rule{4.pt}{0ex}}& =& \phantom{\rule{4.pt}{0ex}}\text{area}\phantom{\rule{4.pt}{0ex}}▵\text{CED}\hfill \\ \hfill \therefore DE\parallel BC& & \phantom{\rule{4.pt}{0ex}}\text{(same}\phantom{\rule{4.pt}{0ex}}\text{side}\phantom{\rule{4.pt}{0ex}}\text{of}\phantom{\rule{4.pt}{0ex}}\text{equal}\phantom{\rule{4.pt}{0ex}}\text{base}\phantom{\rule{4.pt}{0ex}}\text{DE,}\phantom{\rule{4.pt}{0ex}}\text{same}\phantom{\rule{4.pt}{0ex}}\text{area)}\hfill \\ \hfill \therefore A\stackrel{^}{D}E& =& A\stackrel{^}{B}C\phantom{\rule{4.pt}{0ex}}\text{(corres}\phantom{\rule{4.pt}{0ex}}\angle \text{'s)}\hfill \\ \hfill \text{and}\phantom{\rule{4.pt}{0ex}}A\stackrel{^}{E}D& =& A\stackrel{^}{C}B\hfill \end{array}$
$\therefore ▵\text{ADE}\phantom{\rule{4.pt}{0ex}}\text{and}\phantom{\rule{4.pt}{0ex}}▵\text{ABC}\phantom{\rule{4.pt}{0ex}}\text{are}\phantom{\rule{4.pt}{0ex}}\text{equiangular}$
$\therefore ▵ADE\phantom{\rule{0.277778em}{0ex}}|||\phantom{\rule{0.277778em}{0ex}}▵ABC\phantom{\rule{4.pt}{0ex}}\text{(AAA)}$

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

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

Proof :

$\begin{array}{ccc}\hfill \mathrm{Let}\phantom{\rule{3pt}{0ex}}\stackrel{^}{C}& =& x\hfill \\ \hfill \therefore \phantom{\rule{3pt}{0ex}}D\stackrel{^}{A}C& =& {90}^{\circ }-x\phantom{\rule{3pt}{0ex}}\left(\angle \mathrm{\text{'}s of a}\phantom{\rule{2pt}{0ex}}▵\right)\hfill \\ \hfill \therefore \phantom{\rule{3pt}{0ex}}D\stackrel{^}{A}B& =& x\hfill \\ \hfill A\stackrel{^}{B}D& =& {90}^{\circ }-x\phantom{\rule{3pt}{0ex}}\left(\angle \mathrm{\text{'}s of a}\phantom{\rule{2pt}{0ex}}▵\right)\hfill \\ \hfill B\stackrel{^}{D}A& =& C\stackrel{^}{D}A=\stackrel{^}{A}={90}^{\circ }\hfill \end{array}$
$\therefore ▵\mathrm{ABD}|||▵\mathrm{CBA}\phantom{\rule{3pt}{0ex}}\mathrm{and}▵\mathrm{CAD}|||▵\mathrm{CBA}\left(\mathrm{AAA}\right)$
$\therefore \frac{AB}{CB}=\frac{BD}{BA}=\left(\frac{AD}{CA}\right)\mathrm{and}\frac{CA}{CB}=\frac{CD}{CA}=\left(\frac{AD}{BA}\right)$
$\therefore A{B}^{2}=CB×BD\phantom{\rule{3pt}{0ex}}\mathrm{and}\phantom{\rule{3pt}{0ex}}A{C}^{2}=CB×CD$
$\begin{array}{ccc}\hfill \therefore A{B}^{2}+A{C}^{2}& =& CB\left(BD+CD\right)\hfill \\ & =& CB\left(CB\right)\hfill \\ & =& C{B}^{2}\hfill \\ \hfill \mathrm{i}.\mathrm{e}.B{C}^{2}& =& A{B}^{2}+A{C}^{2}\hfill \end{array}$

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

1. $\begin{array}{ccc}\hfill L\stackrel{^}{I}J& =& G\stackrel{^}{I}H\hfill \\ \hfill J\stackrel{^}{L}I& =& H\stackrel{^}{G}I\left(\mathrm{Corres}.\angle \mathrm{s}\right)\hfill \\ \hfill \therefore ▵LIJ& |||& ▵GIH\left(\mathrm{Equiangular}▵\mathrm{s}\right)\hfill \end{array}$
$\begin{array}{ccc}\hfill L\stackrel{^}{I}K& =& G\stackrel{^}{I}J\hfill \\ \hfill K\stackrel{^}{L}I& =& J\stackrel{^}{G}I\left(\mathrm{Corres}.\angle \mathrm{s}\right)\hfill \\ \hfill \therefore ▵LIK& |||& ▵GIJ\left(\mathrm{Equiangular}▵\mathrm{s}\right)\hfill \end{array}$
2. $\begin{array}{ccc}\hfill \frac{HJ}{JI}& =& \frac{GL}{LI}\left(▵LIJ\phantom{\rule{0.277778em}{0ex}}|||\phantom{\rule{0.277778em}{0ex}}▵GIH\right)\hfill \\ \hfill \text{and}\phantom{\rule{4.pt}{0ex}}\frac{GL}{LI}& =& \frac{JK}{KI}\left(▵LIK\phantom{\rule{0.277778em}{0ex}}|||\phantom{\rule{0.277778em}{0ex}}▵GIJ\right)\hfill \\ & =& \frac{5}{3}\hfill \\ \hfill \therefore \frac{HJ}{JI}& =& \frac{5}{3}\hfill \end{array}$
3. $\begin{array}{ccc}\hfill \frac{HJ}{KI}& =& \frac{HJ}{JI}×\frac{JI}{KI}\hfill \end{array}$

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:

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

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

1. $\begin{array}{ccc}\hfill \stackrel{^}{{P}_{1}}& =& \stackrel{^}{{S}_{1}}\left(\mathrm{Alt}.\angle \mathrm{s}\right)\hfill \\ \hfill \stackrel{^}{{Q}_{1}}& =& \stackrel{^}{{R}_{1}}\left(\mathrm{Alt}.\angle \mathrm{s}\right)\hfill \\ \hfill \therefore ▵PTQ& |||& ▵STR\left(\mathrm{Equiangular}▵\mathrm{s}\right)\hfill \end{array}$
2. $\begin{array}{ccc}\hfill \frac{PT}{TQ}& =& \frac{ST}{TR}\left(▵PTQ|||▵STR\right)\hfill \\ \hfill \therefore PT·TR& =& ST·TQ\hfill \end{array}$

## 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 $y$ .

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