 # Triangles

## 06 TRIANGLES

Similarity:

Two figures are similar if their shapes are same but size may not be the same.

e.g Tajmahal and model of Tajmahal

Similarity of triangles:

Two triangles Δ ABC and Δ PQR are similar

If    1) Their corresponding angles are equal i.e

2) Their corresponding sides are in proportion i.e.

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$

Theorem 1:Basic proportionality Theorem:(Thale’s Theorem)

If a line is drawn parallel to one side of triangle intersecting the other two sides, then

it divides the two sides in the same ratio.

Given: A triangle ABC in which

, and DE intersects AB at D and AC at E.

$⇒\frac{AB}{AD}=\frac{AC}{AE}$
$⇒\frac{AD+DB}{DB}=\frac{AE+EC}{EC}$
$⇒\frac{AB}{DB}=\frac{AE}{EC}$

THEOREM 2:(Converse of Basic Proportionality Theorem)

If a line divides any two sides of a triangle in the same ratio, then the line must be

Parallel  to the third side.

From (i) and (ii), we get

$\frac{AF}{FC}=\frac{AE}{EC}$
$⇒\frac{AC}{FC}=\frac{AC}{EC}$
$⇒EC=EC$

THEOREM 3(Angle bisector theorem):

The internal bisector of an angle of a triangle divides the opposite side internally in

the ratio of the sides containing the angle.

From (i) and (ii), we get

$\angle 3=\angle 4$

THEOREM 4(Converse of angle bisector theorem):

CONSTRUCTION :Produce BA to E such that AE = AC. Join EC

Now,

$\frac{BD}{DC}=\frac{AB}{AC}$

THEOREM 5:

If three or more parallel lines are intersected by two transversals, prove that the

intercepts made by them on the transversals are proportional.

GIVEN:Three parallel lines l, m, n which are cut by the transversals AB and CD at P, Q, R and E, F, G respectively.

CONSTRUCTION: join  PG.

PROOF:

PROOF:

$FM\parallel PE$

Criteria for similarities of triangles:

THEOREM6:(AAA Similarity Criterion):

If two triangles are equiangular, then they are similar.

PROOF:  We can have   AB=DE,,AB>DE,AB<DE

Case I: When AB = DE.

In this case, we have

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

Hence proved

Take a point P on the line DE and Q on the line DF such that AB = DP and AC = DQ.

Join PQ.

$\therefore \angle DPQ=\angle DEF$

Similarly, we can prove that

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

Hence proved

Case III:When  AB > DE

Take a point P on the line DE produced and Q on the line DF produced

so  that DP = AB and DQ = AC.

Join PQ.

$\angle D=\angle DPQ$
$\frac{AB}{DE}=\frac{AC}{DF}$

Similarly, we can prove that

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

hence proved

THEOREM 7: (SSS Similarity Criterion)

If the corresponding sides of two triangles are proportional, then they are similar.

CONSTRUCTION:Let P and Q be points on DE and DF respectively

such that DP = AB and DQ = AC.

Join PQ.

PROOF :We have,

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

Thus, in triangles DPQ and DEF, we have

From (i) and (ii), we have

$⇒△ABC\parallel △DEF$

THEOREM 8(SAS similarity Criterion):

If in two triangles, one pair of corresponding sides are proportional and the

included angles are equal then the two triangles are similar.

CONSTRUCTION: Take points P and Q on DE and DF respectively,so that

DP = AB and DQ = AC. Join PQ.

…(ii), (by AAA-criterion of similarity)

From (i) and (ii), we get

Areas of similar triangles:

Theorem 9:

The ratio of areas of two similar triangles is equal to the square of the ratio of their

Corresponding sides

Given: Triangle ABC is similar to triangle PQR.

$\angle B=\angle Q$
$\angle C=\angle R$
$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}$

Proof:

…. (i)

$\frac{A\left(△ABC\right)}{A\left(△PQR\right)}=\frac{BC×AM}{QR×PN}=\frac{BC×BC}{QR×QR}={\left(\frac{BC}{QR}\right)}^{2}$

Similarly we can show for other side too.

$\frac{A\left(△ABC\right)}{A\left(△PQR\right)}={\left(\frac{AB}{PQ}\right)}^{2}={\left(\frac{BC}{QR}\right)}^{2}=\left(\frac{AC}{PR}\right)2$

Theorem 10:Theorem of three right angles triangles

If a perpendicular is drawn from the vertex of the right angle of a right angle

triangle to the hypotenuse then triangles on both sides of perpendicular are

similar to the whole triangle and each other.

Proof:

Theorem 11: Pythagoras theorem:

In a right angle triangle,the square of the hypotenuse is equal to the sum of

squares of other two sides.

$\frac{AB}{AD}=\frac{BC}{BD}=\frac{AC}{AB}$
$⇒\frac{AB}{AD}=\frac{AC}{AB}$
$A{B}^{2}=AD×AC....\left(1\right)$
$\frac{AB}{BD}=\frac{BC}{DC}=\frac{AC}{BC}$
$⇒\frac{BC}{DC}=\frac{AC}{BC}$
$=A{C}^{2}$
$A{B}^{2}+B{C}^{2}=CD×AC+AD×AC$

Hence proved.

Theorem 12: Converse of Pythagoras theorem:

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

Proof: In PQR we can apply Pythagoras theorem

So AC = PR (from 1 and 2)

AC = PR

PQ = AB,

QR = BC

Proved..