## Similarity Maths Notes

In previous standard, we studied that area of any triangle = \(\frac{1}{2}\) × base × height.

In the given figure, seg BC is the base and seg AM is the height of the triangle.

∴ A (∆ ABC) = \(\frac{1}{2}\) × BC × AM

**Ratio of areas of two triangles:**

Statement: Ratio of areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.

Given: In ∆ ABC, seg AM ⊥ side BC such that B-M-C and seg XN ⊥ side YZ such that Y-N-Z.

Proof: Area of any triangle = \(\frac{1}{2}\) × base × height

∴ A(∆ ABC) = \(\frac{1}{2}\) × BC × AM …… (1)

and A(∆ XYZ) = \(\frac{1}{2}\) × YZ × XN … (2)

If the base, height and area of one triangle is denoted by b_{1}, h_{1} and A_{1} respectively and that of the other triangle be denoted by b_{2}, h_{2} and A_{2} respectively then \(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{b_{1} \times h_{1}}{b_{2} \times h_{2}}\)

Consider the following case: Both the triangles are of equal heights.

Thus for two triangles if h_{1} = h<_{2}, then

\(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{b_{1}}{b_{2}}\)

Thus areas of triangles with equal heights are proportional to their corresponding bases.

Consider the following case: Both the triangles are of equal bases.

Thus for two triangles, if b_{1} = b_{2}, then

\(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\frac{h_{1}}{h_{2}}\)

Thus areas of triangles with equal bases are proportional to their corresponding heights.

**Remember this!**

- Ratio of the areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.
- Areas of triangles with equal heights are proportional to their corresponding bases.
- Areas of triangles with equal bases are proportional to their corresponding heights.

**Basic proportionality theorem:**

Statement: If a line parallel X to a side of a triangle, intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.

Given: In ∆ XYZ

- line AB || side YZ.
- line AB intersects side XY and side XZ in points A and B respectively such that X-A-Y and X-B-Z.

To prove: \(\frac{\mathrm{XA}}{\mathrm{AY}}=\frac{\mathrm{XB}}{\mathrm{BZ}}\)

Construction: Draw seg BY and seg AZ.

Proof: ∆ XAB and ∆ BAY have a common vertex B and their bases XA and AY lie on the same line XY.

∴ they have equal heights.

∴ \(\frac{\mathrm{A}(\triangle \mathrm{XAB})}{\mathrm{A}(\triangle \mathrm{BAY})}=\frac{\mathrm{XA}}{\mathrm{AY}}\) ……. [Triangle of equal heights] …. (1)

∆ XAB and ∆ ABZ have a common vertex A and their bases XB and BZ lie on the same line XZ.

∴ they have equal heights.

∆ BAY and ∆ ABZ lie between the same two parallel lines AB and YZ.

∴ they have equal heights, also they have same base AB.

∴ A(∆ BAY) = A(∆ ABZ)

… [Triangles of same base and heights] … (3)

∴ from (1), (2) and (3), we get

\(\frac{\mathrm{A}(\triangle \mathrm{XAB})}{\mathrm{A}(\triangle \mathrm{BAY})}=\frac{\mathrm{A}(\triangle \mathrm{XAB})}{\mathrm{A}(\triangle \mathrm{ABZ})}\)

∴ from (1), (2) and (4), we get

\(\frac{\mathrm{XA}}{\mathrm{AY}}=\frac{\mathrm{XB}}{\mathrm{BZ}}\)

**Converse of basic proportionality theorem:**

Statement: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

In the figure, line XY intersects side PQ in point X and side PR in point Y such that P-X-Q and P-Y-R.

If \(\frac{\mathrm{PX}}{\mathrm{XQ}}=\frac{\mathrm{PY}}{\mathrm{YR}}\), then line XY || side QR.

**Theorem of an angle bisector of a triangle:**

Statement: The bisector of an angle of a triangle divides the side opposite to the angle in tbe ratio of the remaining sides.

Given: In PQR, bisector of ∠PQR intersects side PR in point S such that P-S-R.

To prove : \(\frac{\mathrm{PQ}}{\mathrm{QR}}=\frac{\mathrm{PS}}{\mathrm{SR}}\)

Construction: Draw a line parallel to ray QS through point R. Extend seg PQ to intersect the line at point M such that P-Q-M.

Proof: In ∆PMR,

seg QS || side MR …… (Construction)

∴ by basic proportionality theorem,

\(\frac{\mathrm{PQ}}{\mathrm{QM}}=\frac{\mathrm{PS}}{\mathrm{SR}}\) …….. (1)

Ray QS || side MR and line PM is the transversal.

∴ ∠PQS ≅∠QMR … [Corresponding angles] … (2)

Ray QS || side MR and line QR is the transversal.

∴ ∠ SQR ≅ ∠ QRM … [Alternate angles] … (3)

∠PQS ≅ ∠SQR … [Ray QS is the bisector of ∠ PQR] … (4)

In ∆ QRM,

∠ QMR ≅ ∠ QRM … [From (2), (3) and (4)]

∴ segQR ≅ segQM

… [Converse of isosceles triangle theorem] … (5)

∴ \(\frac{\mathrm{PQ}}{\mathrm{QR}}=\frac{\mathrm{PS}}{\mathrm{SR}}\) …… [From (1) and (5)]

**converse of angle bisector theorem:**

Consider ∆ PQR, ray PS intersects side QR in point S such that Q-S-R.

If \(\frac{\mathrm{PQ}}{\mathrm{PR}}=\frac{\mathrm{QS}}{\mathrm{SR}}\) then ray PS is the bisector of ∠QPR.

**Angle bisector theorem:**

Every point on the bisector of an angle is equidistant from the sides of the angle.

In the figure, point A lies on the bisector of ∠XYZ and seg AB ⊥ ray YX, and seg AC ⊥ ray YZ.

Point A is , equidistant from ray YX and ray YZ.

∴ AB = AC.

**Theorem of an angle bisector of a triangle:**

Alternative method:

Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.

Given: In ∆ ABC, seg AD bisects ∠ BAC such that B-D-C.

To prove: \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}}\)

Construction: Draw seg DM ⊥ side AB such that A-M-B and seg DN ⊥ side AC such that A-N-C.

Proof: Point D lies on the bisector of ∠ BAC

∴ by angle bisector theorem, DM = DN … (1)

Now, the ratio of areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.

Also, ∆ ADB and ∆ ADC have a common vertex A and their bases BD and DC 11e on the same line BC.

∴ their heights are equal.

Areas of triangles with equal heights are proportional to their corresponding bases.

∴ from (2) and (3), we get,

\(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{DC}}\).

**Property of three parallel lines and their transversals:**

Statement: The ratio of the intercepts made on a transversal by three parallel lines is equal to the ratio of the corresponding intercepts made on any other transversal by the same parallel lines.

Given : Line a || line b || line c. Transversal p intersects lines a, b and c in points L, M and N respectively. Transversal q intersects lines a, b and e in points X, Y and Z respectively.

To prove : = \(\frac{\mathrm{LM}}{\mathrm{MN}}=\frac{\mathrm{XY}}{\mathrm{YZ}}\)

Construction : Draw seg LZ. Let seg LZ intersect seg MY at point K such that M-K-Y and L-K-Z.

Proof: In ∆ LNZ.

seg MK || side NZ … (Given)

∴ by basic proportionality theorem,

\(\frac{\mathrm{LM}}{\mathrm{MN}}=\frac{\mathrm{LK}}{\mathrm{KZ}}\) ………. (1)

In ∆LZX

seg KY || side LX .. (Given)

∴ by basic proportionality theorem,

\(\frac{\mathrm{LK}}{\mathrm{KZ}}=\frac{\mathrm{XY}}{\mathrm{YZ}}\) ………. (2)

∴ from (1) and (2),

\(\frac{\mathrm{LM}}{\mathrm{MN}}=\frac{\mathrm{XY}}{\mathrm{YZ}}\)

**Remember this!**

(1) Basic proportionality theorem.

In ∆ ABC,

if seg XY || side BC

then \(\frac{\mathrm{AX}}{\mathrm{XB}}=\frac{\mathrm{AY}}{\mathrm{YC}}\)

(2) Converse of basic proportionality theorem.

In ∆ XYZ, if = \(\frac{\mathrm{XA}}{\mathrm{AY}}=\frac{\mathrm{XB}}{\mathrm{BZ}}\)

then seg AB || side YZ.

(3) Property of an angle bisector of a triangle.

In ∆ PQR, if ray QS bisects ∠ PQR then = \(\frac{\mathrm{PQ}}{\mathrm{QR}}=\frac{\mathrm{PS}}{\mathrm{SR}}\)

(4) Converse of angle bisector theorem.

In ∆ XYZ, if \(\frac{\mathrm{XY}}{\mathrm{YZ}}=\frac{\mathrm{XM}}{\mathrm{MZ}}\) then ray YM bisects ∠XYZ.

(5) Property of three parallel lines and their transversals.

If line AB || line PQ line MN, x and y are the transversals, then = \(\frac{\mathrm{AP}}{\mathrm{PM}}=\frac{\mathrm{BQ}}{\mathrm{QN}}\)

**Similar figures:**

Figures which are of same shape but may be of different sizes are called similar figures.

- It means similar figures are always of same shape but can be of different sizes.
- If similar figures are of same shape and size then those figures are also congruent.

similar triangles: For a given one to one correspondence between the vertices of two triangles, if

(i) their corresponding angles are congruent and

(ii) their corresponding sides are in same proportion, then the correspondence is known as similarity and the two triangles are said to be similar triangles.

In ∆ ABC, if ∠A ≅∠P, ∠B ≅ ∠Q and ∠C ≅ ∠R and

\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AC}}{\mathrm{PR}}\); then ∆ ABC and ∆PQR are similar

i.e. ∆ ABC ~ ∆ PQR under ABC ↔ PQR.

**Test of similarity of triangles:**

To prove two triangles similar, it is not necessary that we need to prove all pairs of corresponding angles congruent and corresponding sides in proportion. Similarity of triangles can be confirmed by testing three specific conditions, often referred as ‘Tests of similarity’. When three specific conditions are fulfilled, the remaining conditions are also fulfilled. There are three tests of similarity of triangles.

- AAA test of similarity or AA test of similarity
- SAS test of similarity
- SSS test of similarity.

**Tests of similarity of triangles:**

(1) AAA test ; For a given one-to-one correspondence between the vertices of two triangles, if the corresponding angles are congruent, then the two triangles are similar.

In the figures, under the’ correspondence, ABC ↔ PQR, ∠A ≅ ∠P, ∠ B ≅∠ Q, ∠ C ≅∠R.

Then by AAA test, ∆ ABC ~ ∆ PQR.

AA test: For a given one-to-one correspondence between the vertices of two triangles, if two angles of one triangle are congruent with the corresponding two angles of the other triangle, then the two triangles are similar.

In the figures, under the correspondence

LMN ↔ RST,

∠M ≅ ∠S, ∠N ≅ ∠T.

Then by AA test, ∆ LMN ~ ∆ RST.

(2) SAS test : For a given one-to-one correspondence between the vertices of two triangles, if two sides of one triangle are proportional to the corresponding sides of the other triangle and the angles included by them are congruent, then the two triangles are similar.

In the figures, under the correspondence

PQR ↔ XYZ,

\(\frac{\mathrm{PQ}}{\mathrm{XY}}=\frac{\mathrm{QR}}{\mathrm{YZ}}=\frac{2}{3}\) and ∠Q ≅ ∠Y. Then by SAS test, ∆PQR ~ ∆XYZ.

(3) sss test : For a given one-to-one correspondence between the vertices of two triangles, if three sides of one triangle are proportional to the three corresponding sides of the other triangle, then the two triangles are similar.

In the figures, under the correspondence

ABC ↔ GEF,

\(\frac{\mathrm{AB}}{\mathrm{GE}}=\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{\mathrm{AC}}{\mathrm{GF}}=\frac{5}{3}\)

Then by SSS test,

∆ABC ~ ∆GFF.

**Properties of Similar Triangles**

1. Reflexivity: Every triangle is similar to itself. Thus, ∆ABC ~ ∆ABC.

2. Symmetry: If one triangle is similar to a second triangle, then the second triangle is similar to the first.

Thus, if ∆ ABC ~ ∆ DEF, then

∆ DEF ~ ∆ ABC.

3. Transitivity : If one triangle is similar to a second triangle and the second triangle is similar to a third triangle, then the first triangle is similar to the third triangle.

Thus, if ∆ ABC ~ ∆ DEF and ∆ DEF ~ ∆ XYZ, then ∆ ABC ~ ∆ XYZ.

**For More Information**

Proof of AAA test:

Given: In ∆ PQR and ∆ XYZ, ∠P ≅ ∠X, ∠Q ≅ ∠Y and ∠R ≅ ∠Z.

T0 prove: ∆ PQR ~ ∆ XYZ

proof: Let us consider ∆ PQR bigger than ∆ XYZ. Mark a point S on side PQ such that XY = PS and P-S-Q. Mark a point T on side PR such that XZ = PT and P-T-R.

In ∆ SPT and ∆ YXZ,

**Theorem of areas of similar triangles:**

Statement: When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides.

Given : ∆ ABC ~ ∆ XYZ

To prove: \(\frac{\mathrm{A}(\triangle \mathrm{ABC})}{\mathrm{A}(\triangle \mathrm{XYZ})}=\frac{\mathrm{AB}^{2}}{\mathrm{XY}^{2}}=\frac{\mathrm{BC}^{2}}{\mathrm{YZ}^{2}}=\frac{\mathrm{AC}^{2}}{\mathrm{XZ}^{2}}\)

Construction : Draw seg AD ⊥ side BC such that

B-D-C and seg XT ⊥ side YZ such that Y-T-Z.

Proof: Ratio of areas of two triangles is equal to the ratio of the products of their bases and corresponding heights.