Similar Triangles

$$\begin{gathered} \Delta \mathrm{ABC}\sim △\mathrm{PQR}(\sim \text{ is the sign of similarity }) \\ \text{ Also, }\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{CA}}{\mathrm{PR}} \end{gathered}$$

Similar figures Two figures are similar if these two conditions are true : (i) Corresponding angles are congruent. (ii) The lengths of corresponding sides are in proportion.

• A proportion is an equation that shows two ratios are equal.

The proportions $\frac{a}{b}=\frac{c}{d}$ and $\mathrm{a}:\mathrm{b}=\mathrm{c}:\mathrm{d}$ can be read as "a is to b as c is to d ".

• The symbol ~ means "is similar to."
  
  PQRS ~ TUVW
• When you name similar figures, be sure to name corresponding vertices in the same order. For example, if PQRS ~ TUVW, then : (i) $\angle \mathrm{P}\cong \angle \mathrm{T},\angle \mathrm{Q}\cong \angle \mathrm{U},\angle \mathrm{R}\cong \angle \mathrm{V}$ and $\angle \mathrm{S}\cong \angle \mathrm{W}$ (ii) $\frac{\mathrm{PQ}}{\mathrm{TU}}=\frac{\mathrm{QR}}{\mathrm{UV}}=\frac{\mathrm{RS}}{\mathrm{VW}}=\frac{\mathrm{SP}}{\mathrm{WT}}$

Numerical Ability 1 Tell whether the shapes in each pair are similar.

Building Concepts 1 Graphic Design : The hiking symbol in the scale drawing is $3$ inch wide and 2.5 inch tall. The symbol will be 10 inch tall on the sign. Find the width of the symbol on the sign. Explanation: Let $\mathrm{w}=$ width of symbol in inches. Set up a proportion. $\frac{\text{ height of symbol in drawing }}{\text{ height of symbol on sign }}=\frac{\text{ width of symbol in drawing }}{\text{ width of symbol on sign }}$ $$\begin{gathered} \frac{2.5}{10}=\frac{3}{\mathrm{w}} \\ \Rightarrow 2.5\mathrm{w}=10\times 3 \\ \mathrm{w}=\frac{30}{2.5} \\ \Rightarrow \mathrm{w}=12 \end{gathered}$$ The hiking symbol will be 12 inch. wide on the sign.

• Two figures are congruent if they have same shape & size, whereas two figures are called similar if they just have same shape, they may differ in size.

Numerical Ability 2 Tell whether the two triangles are similar.

Similar triangles You know several shortcuts for proving that two triangles are congruent without checking each angle and side. In this lesson you will learn some shortcuts for proving that triangles are similar. When two triangles have two pairs of congruent angles, the third pair must be congruent too. If you experiment with two triangles in which all three pairs of angles are congruent, you will find that the lengths of their sides are in proportion. Angle-Angle (AA) similarity postulate If two angles of a triangle are congruent to two angles of another triangle, then the triangles are similar.

• All congruent triangles are similar but all similar triangles are not congruent.

Building Concepts 2 In the following figure $\overline{\mathbf{QT}}\Vert \overline{\mathrm{RS}}$. Prove that $△\mathbf{PRS}\sim \Delta \mathrm{PQT}$

Side-Angle-Side (SAS) similarity theorem

If an angle of one triangle is congruent to an angle of another triangle, and the sides including these angles are in proportion, then the triangles are similar. If $\angle \mathrm{A}\cong \angle \mathrm{D}$ and $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AC}}{\mathrm{DF}}$, (see figure)

Side-Side-Side (SSS) similarity theorem

If all corresponding sides of two triangles are in proportion, then the triangles are similar.

If $\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$, (see figure) then $△\mathrm{ABC}\sim △\mathrm{DEF}$.

Building Concepts 3 Is it possible to prove that the triangles in each pair are similar? Explain why or why not. (i)

Building Concepts 4 Tell whether the triangles in each pair are similar. Explain with reason.

1.0Proportions and similarity

You know that in similar figure the lengths of the sides are in proportion. You can rewrite these proportions in several ways to help you learn more about similar figures. In the Exploration, you will investigate one way that parallel lines and similar figures are related.

Finding proportions in triangles Work with another student. You will need : Geometry software, or patty paper and a ruler

Exploring the concept (i) Draw a scalene triangle, $△\mathrm{PQR}$. Construct $\overline{\mathrm{ST}}$ so that $\overline{\mathrm{ST}}\Vert \overline{\mathrm{PQ}}$ and $\overline{\mathrm{ST}},\overline{\mathrm{RT}}$ and $\overline{\mathrm{RS}}$ form a smaller triangle, $△$ RST. Find the lengths of the sides of the triangles.

Drawing conclusion In the exploration, you saw that a line that intersects two sides of a triangle and is parallel to the third side forms a triangle that is similar to the original one.

Triangle proportionality theorem

If a segment is parallel to one side of a triangle and intersects the other two sides in distinct points, then it divides those sides proportionally.

If $\overline{ST}\Vert \overline{QR}$, then $\frac{PS}{SQ}=\frac{PT}{TR}$. (see figure)

Theorem (Thales theorem or basic proportionality theorem):

If a line is drawn parallel to one side of a triangle intersecting the other two sides, then the other two sides are divided in the same ratio.

Given : $\mathrm{A}△\mathrm{ABC}$ in which line $\ell$ parallel to $\mathrm{BC}(\mathrm{DE}|\mid \mathrm{BC})$ intersecting AB at D and AC at E . To prove : $\frac{AD}{DB}=\frac{AE}{EC}$ Construction : Join D to C and E to B . Through E draw EF perpendicular to AB i.e., $\mathrm{EF}\perp \mathrm{AB}$ and through D draw $\mathrm{DG}\perp \mathrm{AC}$.

Proof: Area of $(△ADE)=\frac{1}{2}(AD\times EF)$ (Area of $\Delta =\frac{1}{2}$ base $\times$ altitude) Area of $(△\mathrm{BDE})=\frac{1}{2}(\mathrm{BD}\times \mathrm{EF})$ Dividing (1) by (2) $\frac{\text{ Area }(△\mathrm{ADE})}{\text{ Area }(△\mathrm{BDE})}=\frac{\frac{1}{2}\mathrm{AD}\times \mathrm{EF}}{\frac{1}{2}\mathrm{BD}\times \mathrm{EF}}=\frac{\mathrm{AD}}{\mathrm{DB}}$

Similarly, $\frac{\text{ Area }(△\mathrm{ADE})}{\text{ Area }(△\mathrm{CDE})}=\frac{\frac{1}{2}\mathrm{AE}\times \mathrm{DG}}{\frac{1}{2}\mathrm{EC}\times \mathrm{DG}}=\frac{\mathrm{AE}}{\mathrm{EC}}$ $\frac{\mathrm{Area}(△\mathrm{ADE})}{\text{ Area }(△\mathrm{CDE})}=\frac{\mathrm{AE}}{\mathrm{EC}}$ Area $(△\mathrm{BDE})=$ Area ( $△\mathrm{CDE}$ ) [ $\Delta \mathrm{s}$ BDE and CDE are on the same base DE and between the same parallel lines DE and BC.]

From (4) and (5) $\frac{\mathrm{Area}(△\mathrm{ADE})}{\mathrm{Area}(△\mathrm{BDE})}=\frac{\mathrm{AE}}{\mathrm{EC}}$ From (3) and (6) $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$ Hence proved.

Theorem (Converse of basic proportionality theorem) : If a line divides any two sides of a triangle proportionally, the line is parallel to the third side.

Given : $△ABC$ and $DE$ is a line meeting $AB$ and $AC$ at $D$ and E respectively such that $\frac{AD}{DB}=\frac{AE}{EC}$ To prove: $\mathrm{DE}|\mid \mathrm{BC}$ Proof: If possible, let DE be not parallel to BC. Then, draw DF || BC $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AF}}{\mathrm{FC}}$ ... (1) (By Basic Proportionality theorem.) $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$ ... (2) (Given)

From (1) and (2) $\therefore \frac{\mathrm{AF}}{\mathrm{FC}}=\frac{\mathrm{AE}}{\mathrm{EC}}$

Adding 1 on both the sides, $\Rightarrow \frac{\mathrm{AF}}{\mathrm{FC}}+1=\frac{\mathrm{AE}}{\mathrm{EC}}+1$ $\Rightarrow \frac{\mathrm{AF}+\mathrm{FC}}{\mathrm{FC}}=\frac{\mathrm{AE}+\mathrm{EC}}{\mathrm{EC}}$ $\Rightarrow \frac{\mathrm{AC}}{\mathrm{FC}}=\frac{\mathrm{AC}}{\mathrm{EC}}(\because \mathrm{AF}+\mathrm{FC}=\mathrm{AC}$ and $\mathrm{AE}+\mathrm{EC}=\mathrm{AC}$. $\Rightarrow \mathrm{FC}=\mathrm{EC}$ $\Rightarrow E$ and $F$ coincide. But, DF||BC. Hence, DE||BC. Hence proved.

Numerical Ability 3 Find the length JN.

• Proportions Two equations are equivalent if one equation can be changed into the other using algebra. For example, $\frac{a}{b}=\frac{c}{d}$ is equivalent to $ad=bc$.
• Properties of proportions All of the proportions below are equivalent to each other. $\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{c}}{\mathrm{d}}\iff \frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{d}}{\mathrm{c}}\iff \frac{\mathrm{a}}{\mathrm{c}}=\frac{\mathrm{b}}{\mathrm{d}}\iff \frac{\mathrm{c}}{\mathrm{a}}=\frac{\mathrm{d}}{\mathrm{b}}$ These proportions are also equivalent to the ones above. $\frac{a}{a+b}=\frac{c}{c+d}$ $\frac{b}{a+b}=\frac{d}{c+d}$ $\frac{\mathrm{a}+\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{c}+\mathrm{d}}{\mathrm{c}}$

If one angle of an isosceles triangle is congruent to one angle of another isosceles triangle, then the triangles are similar.

Equivalent proportions can be used to prove the Triangle proportionality theorem. Given: $\overline{\mathrm{ST}}\Vert \overline{\mathrm{QR}}$ (see figure) To prove : $\frac{PS}{SQ}=\frac{PT}{TR}$ Proof: $\overline{\mathrm{ST}}\Vert \overline{\mathrm{QR}}$

Numerical Ability 4 Find each value. (i) $\frac{\mathrm{LK}}{\mathrm{KJ}}$ (ii) $\frac{\mathrm{KN}}{\mathrm{LM}}$

2.0Results on area of similar triangles

Theorem : The areas of two similar triangles are proportional to the squares of their corresponding sides. Given : $△\mathrm{ABC}\sim △\mathrm{DEF}$ To prove: $\frac{\text{ Area of }△\mathrm{ABC}}{\text{ Area of }△\mathrm{DEF}}=\frac{{\mathrm{AB}}^2}{{\mathrm{DE}}^2}=\frac{{\mathrm{BC}}^2}{{\mathrm{EF}}^2}=\frac{{\mathrm{AC}}^2}{{\mathrm{DF}}^2}$ Construction : Draw AL $\perp$ BC and $\mathrm{DM}\perp \mathrm{EF}$. Proof: $\frac{\text{ Area of }△\mathrm{ABC}}{\text{ Area of }△\mathrm{DEF}}=\frac{\frac{1}{2}\times \mathrm{BC}\times \mathrm{AL}}{\frac{1}{2}\times \mathrm{EF}\times \mathrm{DM}}$ (Area of $\Delta =\frac{1}{2}\times$ Base $\times$ Height) $\Rightarrow \frac{\text{ Area of }△ABC}{\text{ Area of }△DEF}=\frac{BC}{EF}\times \frac{AL}{DM}$ In $△\mathrm{ALB}$ and $△\mathrm{DME}$, we have $\angle \mathrm{ALB}=\angle \mathrm{DME}$ (Each equal to $90^{\circ }$ ) $\angle \mathrm{ABL}=\angle \mathrm{DEM}$ $(△\mathrm{ABC}\sim △\mathrm{DEF}\Rightarrow \angle \mathrm{B}=\angle \mathrm{E})$ $\therefore \Delta \mathrm{ALB}\sim \Delta \mathrm{DME}$ (by AA-axiom) $\Rightarrow \frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\mathrm{AB}}{\mathrm{DE}}$ (Corresponding sides of similar $\Delta \mathrm{s}$ are proportional.) $\Delta \mathrm{ABC}\sim \Delta \mathrm{DEF}$ (Given) $\Rightarrow \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{DF}}$ (Corresponding sides of similar $\Delta \mathrm{s}$ are proportional.) From (2) and (3)

Corollary-1: The areas of two similar triangles are proportional to the squares of their corresponding altitude. Given : $△\mathrm{ABC}\sim \Delta \mathrm{DEF},\mathrm{AL}\perp \mathrm{BC}$ and $\mathrm{DM}\perp \mathrm{EF}$. To prove: $\frac{\text{ Area of }△\mathrm{ABC}}{\text{ Area of }△\mathrm{DEF}}=\frac{{\mathrm{AL}}^2}{{\mathrm{DM}}^2}$ Proof: (Area of $\Delta =\frac{1}{2}\times$ Base $\times$ Height $)$

Corollary-2 : The areas of two similar triangles are proportional to the squares of their corresponding medians. Given : $△\mathrm{ABC}\sim △\mathrm{DEF}$ and $\mathrm{AP},\mathrm{DQ}$ are their medians. To prove: $\frac{\text{ Area of }△ABC}{\text{ Area of }△DEF}=\frac{{A{P}^2}_{D{Q}^2}}{}$ Proof: $△\mathrm{ABC}\sim △\mathrm{DEF}$ (Given) $\Rightarrow \frac{\text{ Area of }△\mathrm{ABC}}{\text{ Area of }△\mathrm{DEF}}=\frac{{\mathrm{AB}}^2}{{\mathrm{DE}}^2}$ (Areas of two similar $\Delta$ s are proportional to the squares of their corresponding sides.) $\Delta \mathrm{ABC}\sim \Delta \mathrm{DEF}$ $\Rightarrow \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{2\mathrm{BP}}{2\mathrm{EQ}}=\frac{\mathrm{BP}}{\mathrm{EQ}}$ (Corresponding sides of similar $\Delta \mathrm{s}$ are proportional.) $\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BP}}{\mathrm{EQ}}$ and $\angle \mathrm{B}=\angle \mathrm{E}$ (From (2) and the fact the $△\mathrm{ABC}\sim △\mathrm{DEF}$ ) $\Rightarrow △\mathrm{APB}\sim △\mathrm{DQE}$ (By SAS-similarity axiom) $\Rightarrow \frac{\mathrm{BP}}{\mathrm{EQ}}=\frac{\mathrm{AP}}{\mathrm{DQ}}$ From (2) and (3) $\Rightarrow \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AP}}{\mathrm{DQ}}\Rightarrow \frac{{\mathrm{AB}}^2}{{\mathrm{DE}}^2}=\frac{{\mathrm{AP}}^2}{{\mathrm{DQ}}^2}$ From (1) and (4) $\frac{\text{ Area of }△\mathrm{ABC}}{\text{ Area of }△\mathrm{DEF}}=\frac{{\mathrm{AP}}^2}{D{Q}^2}$ Hence proved.

Corollary-3 : The areas of two similar triangles are proportional to the squares of their corresponding angle bisector segments. Given : $△\mathrm{ABC}\sim △\mathrm{DEF}$ and $\mathrm{AX},\mathrm{DY}$ are their bisectors of $\angle \mathrm{A}$ and $\angle \mathrm{D}$ respectively. To prove: $\frac{\text{ Area of }△ABC}{\text{ Area of }△DEF}=\frac{A{X}^2}{D{Y}^2}$ Proof: $\frac{\text{ Area of }△ABC}{\text{ Area of }△DEF}=\frac{A{B}^2}{D{E}^2}$ (Areas of two similar $\Delta \mathrm{s}$ are proportional to the squares of the corresponding sides.) $△\mathrm{ABC}\sim \Delta \mathrm{DEF}$ (Given) $\Rightarrow \angle \mathrm{A}=\angle \mathrm{D}$ $\Rightarrow \frac{1}{2}\angle \mathrm{A}=\frac{1}{2}\angle \mathrm{D}$

Numerical Ability 5 It is given that $△ABC\sim △PQR$, area $(△ABC)=36{\mathrm{cm}}^2$ and area $(△PQR)=25{\mathrm{cm}}^2$. If $QR=6\mathrm{cm}$, find the length of $BC$.

Numerical Ability 6 $P$ and $Q$ are points on the sides $AB$ and $AC$ respectively of $△ABC$ such that $PQ\Vert BC$ and divides $△\mathrm{ABC}$ into two parts, equal in area. Find PB : AB .

3.0Pythagoras theorem

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Converse of Pythagoras theorem

In a triangle if the square of one side is equal to the sum of the squares of the other two sides, then the triangle is right angled.

Given : $\mathrm{A}△\mathrm{ABC}$ in which ${\mathrm{AB}}^2+{\mathrm{BC}}^2={\mathrm{AC}}^2$ To prove : $\angle \mathrm{B}=90^{\circ }$ Construction : Draw a $△\mathrm{DEF}$ in which $\mathrm{DE}=\mathrm{AB},\mathrm{EF}=\mathrm{BC}$ and $\angle \mathrm{E}=90^{\circ }$

Proof: In $△DEF$, we have: $\angle E=90^{\circ }$ $\therefore {\mathrm{DE}}^2+{\mathrm{EF}}^2={\mathrm{DF}}^2$ (By Pythagoras Theorem) $\Rightarrow {\mathrm{AB}}^2+{\mathrm{BC}}^2={\mathrm{DF}}^2$ $(\because \mathrm{DE}=\mathrm{AB}$ and $\mathrm{EF}=\mathrm{BC})$ $\Rightarrow {\mathrm{AC}}^2={\mathrm{DF}}^2$ $\left(\because {\mathrm{AB}}^2+{\mathrm{BC}}^2={\mathrm{AC}}^2\right)$ $\Rightarrow \mathrm{AC}=\mathrm{DF}$

In $△\mathrm{ABC}$ and $△\mathrm{DEF}$, we have : $\mathrm{AB}=\mathrm{DE}$ (By construction) $\mathrm{BC}=\mathrm{EF}$ (By construction) $\mathrm{AC}=\mathrm{DF}$ (Proved above) $\therefore △\mathrm{ABC}\cong △\mathrm{DEF}$ (By SSS congruence) $\Rightarrow \angle \mathrm{B}=\angle \mathrm{E}$ (by c.p.c.t) $\Rightarrow \angle B=90^{\circ }$ $\left(\because \angle \mathrm{E}=90^{\circ }\right)$ Hence, $\angle \mathrm{B}=90^{\circ }$

Numerical Ability 7 If $ABC$ is an equilateral triangle of side $a$, prove that its altitude is $\frac{\sqrt{3}}{2}a$. Solution:

Numerical Ability 8 In a $△ABC$, obtuse angled at $B$, if $AD$ is perpendicular to $CB$ produced, prove that : $A{C}^2=A{B}^2+{\mathrm{BC}}^2+2\mathrm{BC}\times \mathrm{BD}$ Solution:

4.0Internal angle bisector theorem

The internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle. If $\angle \mathrm{BAD}=\angle \mathrm{CAD}$ Then, $\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}}$

5.0Memory Map