What is a similar figure? In geometry, similar figures are shapes that have the exact same shape, but not necessarily the same size. Think of a photograph of yourself: whether it is printed as a small passport-size photo or a large wall poster, your features look exactly the same because the proportions are preserved. Only the size changes!
In contrast, congruent figures are identical twins: they have both the same shape and the same size. Therefore, all congruent figures are similar, but all similar figures are not congruent. Two figures having the same shape (and not necessarily the same size) are called similar figures, and the symbol for similarity is $\sim$.
Similarity of polygons. 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. proportional). Both conditions are essential. A square and a rectangle have all angles equal (90$^\circ$ each) but their sides are not proportional, so they are not similar. A square and a rhombus have proportional sides but unequal angles, so again they are not similar. All circles are similar, all squares are similar, and all equilateral triangles are similar.
For triangles the situation is special and very useful: if the corresponding angles of two triangles are equal, then their corresponding sides are automatically proportional, and conversely if the corresponding sides are proportional then the corresponding angles are automatically equal. This is why the criteria below need fewer checks for triangles than for general polygons.
Basic Proportionality Theorem (Thales Theorem). 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. In $\triangle ABC$, if a line parallel to $BC$ meets $AB$ at $D$ and $AC$ at $E$, then $\dfrac{AD}{DB}=\dfrac{AE}{EC}$.
Proof of BPT. Join $BE$ and $CD$, and drop perpendiculars $EM\perp AB$ and $DN\perp AC$. Then $\text{ar}(\triangle ADE)=\tfrac12\,AD\cdot EM$ and $\text{ar}(\triangle BDE)=\tfrac12\,DB\cdot EM$, so $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)}=\dfrac{AD}{DB}$. Similarly $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CDE)}=\dfrac{AE}{EC}$. Now $\triangle BDE$ and $\triangle CDE$ are on the same base $DE$ and between the same parallels $DE$ and $BC$, so they have equal areas. Hence $\dfrac{AD}{DB}=\dfrac{AE}{EC}$.
Converse of BPT. If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. That is, if $\dfrac{AD}{DB}=\dfrac{AE}{EC}$ then $DE\parallel BC$. This converse is the tool we use to prove two lines parallel.
Criteria for Similarity of Triangles. Instead of checking all three angles and all three sides every time, we use short-cut rules:
1. AAA / AA criterion. If in two triangles the corresponding angles are equal, then their corresponding sides are in the same ratio and the triangles are similar. Because the angle sum of a triangle is 180$^\circ$, if two pairs of angles match, the third pair must match too; so two equal angle pairs are enough — this is the AA criterion. 2. SSS criterion. If the corresponding sides of two triangles are in the same ratio, their corresponding angles are equal and the triangles are similar. 3. SAS criterion. If one angle of a triangle equals one angle of another triangle and the sides including these angles are proportional, the triangles are similar.
Writing similarity correctly. When we write $\triangle ABC\sim\triangle PQR$, the order of letters matters: it tells us $\angle A=\angle P,\ \angle B=\angle Q,\ \angle C=\angle R$ and $\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{CA}{RP}$. A very common mistake is to match sides in the wrong order — always read corresponding vertices in the same position.
Common mistakes to avoid. (i) Claiming two figures are similar from equal angles alone (true only for triangles, not general polygons). (ii) In BPT, dividing as $\dfrac{AD}{AB}=\dfrac{AE}{EC}$ instead of using matching parts — keep the same kind of segment top and bottom. (iii) Forgetting that AA needs the angles to be corresponding. (iv) Confusing similarity ($\sim$) with congruence ($\cong$).
| Feature | Congruent Triangles | Similar Triangles |
|---|
| Shape | Exactly the same | Exactly the same |
| Size | Exactly the same | Can be different |
| Corresponding Angles | Equal | Equal |
| Corresponding Sides | Equal (Ratio is 1:1) | Proportional (Ratio is equal) |
| Symbol | ≅ | ~ |
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In $\triangle ABC$ and $\triangle PQR$, $\angle A=50^\circ$, $\angle B=70^\circ$, $\angle P=50^\circ$ and $\angle Q=70^\circ$. Are these triangles similar? State the criterion.
Solution- Step 1: Match the corresponding angles. $\angle A=\angle P=50^\circ$ and $\angle B=\angle Q=70^\circ$.
- Step 2: By the angle sum property the third pair must also be equal: $180^\circ-50^\circ-70^\circ=60^\circ$ in both triangles.
- Step 3: Two pairs of corresponding angles are equal, so the AA criterion applies.
Answer: Yes, $\triangle ABC\sim\triangle PQR$ by the AA similarity criterion.
A 6-foot tall student stands beside a vertical flagpole. The student casts a 4-foot shadow while the flagpole casts a 20-foot shadow. Find the height of the flagpole.
Solution- Step 1: The sun's rays make equal angles with the ground for both objects, and both objects are vertical ($90^\circ$). So the two triangles are similar by AA.
- Step 2: Corresponding sides are proportional: $\dfrac{\text{Height of pole}}{\text{Height of student}}=\dfrac{\text{Shadow of pole}}{\text{Shadow of student}}$.
- Step 3: Let the height be $H$. Then $\dfrac{H}{6}=\dfrac{20}{4}=5$.
- Step 4: So $H=5\times 6=30$ feet.
Answer: The flagpole is 30 feet tall.
In $\triangle ABC$, $DE\parallel BC$ with $D$ on $AB$ and $E$ on $AC$. If $AD=2$ cm, $DB=4$ cm and $AC=9$ cm, find $AE$.
Solution- Step 1: Since $DE\parallel BC$, by the Basic Proportionality Theorem $\dfrac{AD}{AB}=\dfrac{AE}{AC}$.
- Step 2: $AB=AD+DB=2+4=6$ cm.
- Step 3: Substitute: $\dfrac{2}{6}=\dfrac{AE}{9}$, i.e. $\dfrac{1}{3}=\dfrac{AE}{9}$.
- Step 4: So $AE=\dfrac{9}{3}=3$ cm.
Answer: $AE=3$ cm.
In $\triangle ABC$, $DE\parallel BC$. If $AD=x$, $DB=x-2$, $AE=x+2$ and $EC=x-1$, find $x$.
Solution- Step 1: By BPT, $\dfrac{AD}{DB}=\dfrac{AE}{EC}$, so $\dfrac{x}{x-2}=\dfrac{x+2}{x-1}$.
- Step 2: Cross-multiply: $x(x-1)=(x+2)(x-2)$.
- Step 3: Expand: $x^2-x=x^2-4$.
- Step 4: Cancel $x^2$: $-x=-4$, so $x=4$.
Answer: $x=4$.
In $\triangle ABC$, $D$ and $E$ lie on $AB$ and $AC$ with $AD=4$ cm, $DB=4.5$ cm, $AE=8$ cm and $EC=9$ cm. Is $DE\parallel BC$?
Solution- Step 1: Compute $\dfrac{AD}{DB}=\dfrac{4}{4.5}=\dfrac{8}{9}$.
- Step 2: Compute $\dfrac{AE}{EC}=\dfrac{8}{9}$.
- Step 3: Since $\dfrac{AD}{DB}=\dfrac{AE}{EC}$, by the Converse of BPT the line $DE$ is parallel to $BC$.
Answer: Yes, $DE\parallel BC$.
It is given that $\triangle ABC\sim\triangle PQR$ with $AB=4$ cm, $BC=5$ cm, $CA=6$ cm and $PQ=8$ cm. Find $QR$ and $RP$.
Solution- Step 1: From the similarity, $\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{CA}{RP}$.
- Step 2: Scale factor $\dfrac{AB}{PQ}=\dfrac{4}{8}=\dfrac12$.
- Step 3: $\dfrac{BC}{QR}=\dfrac12\Rightarrow QR=2\times 5=10$ cm.
- Step 4: $\dfrac{CA}{RP}=\dfrac12\Rightarrow RP=2\times 6=12$ cm.
Answer: $QR=10$ cm, $RP=12$ cm.
The diagonals of a trapezium $ABCD$ with $AB\parallel DC$ intersect at $O$. Using similar triangles, show that $\dfrac{AO}{OC}=\dfrac{BO}{OD}$.
Solution- Step 1: In $\triangle AOB$ and $\triangle COD$, $\angle AOB=\angle COD$ (vertically opposite angles).
- Step 2: Since $AB\parallel DC$, $\angle OAB=\angle OCD$ and $\angle OBA=\angle ODC$ (alternate angles).
- Step 3: By the AA criterion, $\triangle AOB\sim\triangle COD$.
- Step 4: Corresponding sides are proportional: $\dfrac{AO}{CO}=\dfrac{BO}{DO}$, i.e. $\dfrac{AO}{OC}=\dfrac{BO}{OD}$.
Answer: Proved: $\dfrac{AO}{OC}=\dfrac{BO}{OD}$.
In $\triangle ABC$, the bisector $AD$ of $\angle A$ meets $BC$ at $D$. If $AB=5$ cm, $AC=7$ cm and $BD=3$ cm, find $DC$. (Use the angle-bisector property $\dfrac{BD}{DC}=\dfrac{AB}{AC}$.)
Solution- Step 1: By the internal angle-bisector property, $\dfrac{BD}{DC}=\dfrac{AB}{AC}$.
- Step 2: Substitute: $\dfrac{3}{DC}=\dfrac{5}{7}$.
- Step 3: Cross-multiply: $5\cdot DC=21$.
- Step 4: $DC=\dfrac{21}{5}=4.2$ cm.
Answer: $DC=4.2$ cm.
Diagonal $BD$ of a parallelogram $ABCD$ intersects segment $AE$ at $F$, where $E$ is any point on $BC$. Prove that $DF\cdot EF=FB\cdot FA$.
Solution- Step 1: In $\triangle DFA$ (note $AD\parallel BC$ so $AD\parallel BE$) and $\triangle BFE$, we have $\angle DFA=\angle BFE$ (vertically opposite).
- Step 2: $\angle FDA=\angle FBE$ (alternate angles, $AD\parallel BE$).
- Step 3: By AA, $\triangle DFA\sim\triangle BFE$.
- Step 4: So $\dfrac{DF}{BF}=\dfrac{FA}{FE}$, which gives $DF\cdot FE=BF\cdot FA$.
Answer: Proved: $DF\cdot EF=FB\cdot FA$.
In $\triangle ABC$, $D$ is the midpoint of $AB$ and $E$ is a point on $AC$ such that $DE\parallel BC$. If $AC=10$ cm, find $AE$.
Solution- Step 1: $D$ is the midpoint of $AB$, so $\dfrac{AD}{DB}=1$.
- Step 2: By BPT, $\dfrac{AE}{EC}=\dfrac{AD}{DB}=1$, so $E$ is the midpoint of $AC$.
- Step 3: Hence $AE=\dfrac12\,AC=\dfrac{10}{2}$.
Answer: $AE=5$ cm.
Two poles of heights 6 m and 11 m stand on level ground. If the line joining their tops makes a triangle, prove (using similarity) that the foot of the perpendicular from the shorter top splits the taller pole proportionally, then find the height above ground at which a horizontal line from the shorter top meets the taller pole.
Solution- Step 1: A horizontal line from the top of the 6 m pole meets the taller pole at a height of 6 m above the ground (it is horizontal, so equal heights).
- Step 2: The remaining part of the taller pole above this line is $11-6=5$ m.
- Step 3: This confirms the horizontal cut occurs at 6 m, leaving a 5 m segment at the top.
Answer: The horizontal line meets the taller pole at a height of 6 m, leaving a 5 m segment above it.
In $\triangle ABC$, points $D$ and $E$ are on $AB$ and $AC$ such that $DE\parallel BC$. If $AD=4$ cm, $AB=12$ cm and $AC=15$ cm, find $AE$ and $EC$.
Solution- Step 1: Since $DE\parallel BC$, by BPT $\dfrac{AD}{AB}=\dfrac{AE}{AC}$.
- Step 2: Substitute: $\dfrac{4}{12}=\dfrac{AE}{15}$, i.e. $\dfrac{1}{3}=\dfrac{AE}{15}$.
- Step 3: $AE=\dfrac{15}{3}=5$ cm.
- Step 4: $EC=AC-AE=15-5=10$ cm.
Answer: $AE=5$ cm and $EC=10$ cm.