TRIANGLES-Notes
Concept of Similar Figures
Similarity of Triangles
- Corresponding angles match
- Corresponding sides maintain consistent ratios
Basic Proportionality Theorem (Thales’ Theorem)
In easy words: When you cut a triangle with a line that stays perfectly parallel to one of its sides, the remaining two sides get sliced into equal ratios.
The cuts may not be equal in length, but their ratio remains the same.
Theorem-1
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Proof:
We need to prove that \[\frac{AD}{DB}=\frac{AE}{AC}\]Construction:
Let us join BE and CD and then draw DM \(\perp\) AC and EN \(\perp\) AB.
Area of \(\Delta ADE\ (=\frac{1}{2} \text{ base }\times\text{ height})\) = \(\frac{1}{2}AD\times EN\)
\[ ar(ADE) = \frac{1}{2}AD\times EN \] Similarly, \[ ar(BDE) = \frac{1}{2}BD\times EN \] Also \[ar(ADE) = \frac{1}{2}AE\times DM\]being \(DM\perp AE\)
and \[ar(DEC) = \frac{1}{2}EC\times DM\]being \(DM\perp EC \text{(extended to A)}\)
therfore, \[\begin{align}\frac{ar(ADE)}{ar(BDE)}&=\frac{\frac{1}{2}AD\times EN}{\frac{1}{2}BD\times EN}\\ &=\frac{AD}{DB}\tag{1}\end{align}\] and \[ \begin{align} \frac{ar(ADE)}{ar(DEC)}&=\frac{\frac{1}{2}AE\times DM}{\frac{1}{2}EC\times DM}\\ &=\frac{AE}{AC}\tag{2} \end{align} \]Note that \(\Delta BDE\) and \(\Delta DEC\) are on the same base DE and between the same parallels BC and DE.
therfore, \[ar(BDE)=ar(DEC)\tag{3}\]therefore, (1) and (2) are equal as ar(ADE) is divided by same quantity (3)
\[\boxed{\frac{AD}{DB}=\frac{AE}{AC}}\]Theorem-2
If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Proof:
Let line DE such that \[\frac{AD}{DB}=\frac{AE}{EC}\tag{1}\] let us assume that DE is not parallel to line BC, but line DE' is parallel to BC therefore, \[\frac{AD}{DB}=\frac{AE'}{E'C}\tag{2}\] Also, from (1) \[\frac{AE}{EC}=\frac{AE'}{E'C}\tag{3}\] adding 1 to both side of (3) \[ \begin{align} \frac{AE}{EC}+1&=\frac{AE'}{E'C}+1\\\\ \frac{AE+EC}{EC}&=\frac{AE'+E'C}{E'C}\tag{4} \end{align} \](4) can only be true when points E' and E coincides
hence, \(DE\parallel BC\)Example-1
If a line intersects sides \(AB\) and \(AC\) of a \(\Delta ABC\) at \(D\) and \(E\) respectively and is parallel to BC, prove that \(\frac{AD}{AB}=\frac{AE}{AC}\)
Proof:
Given: Line DE intersect line \(AB\) and \(AC\) at point \(D\) and \(E\) and
$$DE\parallel BC$$ To Prove: $$\dfrac{AD}{AB}=\dfrac{AE}{AC}$$ Proof: $$ DE\parallel BC\quad\text{(Given)}$$ therefore, by BPT (Basic Proportionality theorem.) $$\dfrac{AD}{DB}=\dfrac{AE}{EC}$$ or $$\dfrac{DB}{AD}=\dfrac{EC}{AE}\tag{1}$$ Adding 1 to both side of equation-(1) $$\begin{aligned}\dfrac{DB}{AD}+1&=\dfrac{EC}{AE}+1\\\\ \dfrac{DB+AD}{AD}&=\dfrac{EC+AE}{AE}\\\\ \Rightarrow \dfrac{AB}{AD}&=\dfrac{AC}{AE}\end{aligned}$$ Hence Proved.Example-2
ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB (see Fig. 6.14). Show that \[\frac{AE}{ED}=\frac{BF}{FC}\]
Solution:
Given: $$\begin{aligned} AB\| DC\\ EF\parallel AB\end{aligned}$$ To Prove: $$\dfrac{AE}{ED}=\dfrac{BF}{FC}$$Construction: Join AC
Proof:In \(\Delta ADC\)
$$ EG\parallel DC\quad\text{(As }EF\parallel DC)$$ By BPT $$\dfrac{AE}{ED}=\dfrac{AG}{GC}\tag{1}$$In \(\Delta ABC\)
$$ AB\parallel GF \quad\text{(As } AB\parallel EF)$$ By BPT $$\dfrac{BF}{FC}=\dfrac{AG}{GC}\tag{2}$$ From (1) and (2) $$\dfrac{AE}{AD}=\dfrac{BF}{FC}$$ Hence Proved.Criteria for Similarity of Triangles (AA, SAS, SSS)
- AA (Angle–Angle Similarity):
If two angles of one triangle match the two of another, their similarity is guaranteed. - SAS (Side–Angle–Side Similarity):
When an angle is sandwiched between two proportional sides in both triangles, similarity follows. - SSS (Side–Side–Side Similarity):
When all three corresponding sides share the same ratio, the two triangles must be similar.
Theorem-3
If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.
Proof:
Consider triangles ABC and DEF such that \(\angle A = \angle D, \angle B = \angle E\text{ and }\angle C = \angle F\) (as in Fig. 6.24)
Construction: Cut DP = AB and DQ = AC and join PQ.So, By SAS rule of congruency \[ \begin{aligned}\Delta ABC&\cong \Delta DPQ\\\\ \Rightarrow \angle B=\angle P&=\angle E \text{ (By CPCT)} \end{aligned} \] \[\begin{align}\angle E= \angle P \text{ and }\angle F &= \angle Q \text{ (Corresponding Angles)}\\\\ PQ&\parallel EF\end{align}\] therefore, $$\dfrac{DP}{PE}=\dfrac{DQ}{QF}$$ or $$\begin{aligned}\dfrac{PE}{DP}&=\dfrac{QF}{DQ}\\\\ \dfrac{PE}{DP}+1&=\dfrac{QF}{DQ}+1\\\\ \dfrac{PE+DP}{DP}&=\dfrac{QF+DQ}{DQ}\\\\ \dfrac{DE}{DP}&=\dfrac{DF}{DQ}\\\\ DP=AB\text{ and }&DQ=AC\\\\ \dfrac{DE}{AB}&=\dfrac{DF}{AC}\\\\ \Rightarrow \dfrac{AB}{DE}&=\dfrac{AC}{DF}\end{aligned}$$ Similarly $$\dfrac{AB}{DE}=\dfrac{BC}{EF}$$ and $$\dfrac{AB}{DE}=\dfrac{BC}{EF}=\dfrac{AC}{DF}$$
Theorem-4
If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similiar.
Proof:
Consider two triangles \(ABC\) and \(DEF\) such that \[\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}\] Construction: Cut \(DP = AB\) and \(DQ = AC\) and join \(PQ\) By BPT $$\dfrac{DP}{PE}=\dfrac{DQ}{QF}$$ and $$ PQ\parallel EF$$ So, $$\angle P=\angle E$$ and $$\angle Q=\angle F$$ therefore, by BPT $$\begin{aligned}&\dfrac{DP}{DE}=\dfrac{DQ}{DF}=\dfrac{PQ}{EF}\\\\ &\dfrac{DP}{PE}=\dfrac{DQ}{DF}=\dfrac{BC}{EF}\quad\left( BC=PQ\right) \end{aligned}$$ Thus $$\begin{aligned}\Delta ABC&\cong \triangle DPQ\\\\ \angle A=\angle D,\ \angle B=&\angle E,\ \angle C=\angle F \quad\left( CPCT\right) \end{aligned}$$
Theorem-5
If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.
Proof:
Consider two Triangles \(ABC\) and \(DEF\) such that $$\dfrac{AB}{DE}=\dfrac{AC}{DF}( < 1) $$Construction: Cut \(DP=AB\) and \(DF=AC\) and join \(PQ\)
$$\begin{aligned}\dfrac{AB}{DE}&=\dfrac{AC}{DF}\\ \Rightarrow \dfrac{DP}{DE}&=\dfrac{DQ}{DF}\end{aligned}$$ \[AB=DP,\ AC=DQ\quad\scriptsize\text{ (By Construction)}\] $$\begin{aligned}\Rightarrow PQ\parallel EF\quad\scriptsize\text{ (By BPT)} \end{aligned}$$ $$\begin{aligned}AB&=DP\quad\scriptsize\text{ (By construction)}\\ AC&=DQ\quad\scriptsize\text{ (By construction)}\\ \angle A&=\angle A\quad\scriptsize\text{ (Common Angle)}\\\\ \Delta ABC&\cong \Delta DPQ\scriptsize\text{ (SAS Rule)}\end{aligned}$$ $$\angle A=\angle D,\angle B=\angle P,\angle C=\angle Q$$ (By CPCT) Hence $$\triangle ABC\sim \triangle DEF$$Example-3
In Fig. 6.29, if \(PQ \parallel RS\), prove that \(\Delta POQ \sim \Delta SOR\).
To prove$$\Delta POQ\sim \Delta SOR$$ Proof: Given $$ PQ\parallel RS$$ \[\text{In } \Delta POQ \text{ and } \Delta SOR\] $$\angle P=\angle S\quad\scriptsize\text{(Alternate Angle)}$$ $$\angle Q=\angle R\quad\scriptsize\text{(Alternate Angle)}$$ $$\angle POQ=\angle SOR\scriptsize\text{ (vertically opposite Angles)}$$ $$\therefore \Delta POQ\sim SOR\scriptsize\text{ (AAA similarity Criterion)}$$
Important Points
- Two figures having the same shape but not necessarily the same size are called similar figures.
- All the congruent figures are similar but the converse is not true.
- Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (i.e., proportion).
- If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
- If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
- If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar (AAA similarity criterion).
- If in two triangles, two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are similar (AA similarity criterion).
- If in two triangles, corresponding sides are in the same ratio, then their corresponding angles are equal and hence the triangles are similar (SSS similarity criterion).
- If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio (proportional), then the triangles are similar (SAS similarity criterion).
Recent posts
Share this Chapter
Found this helpful? Share this chapter with your friends and classmates.
💡 Exam Tip: Share helpful notes with your study group. Teaching others is one of the fastest ways to reinforce your own understanding.