## Triangles

- All congruent figures are similar, but it doesn’t mean that all similar figures are congruent.

- Two polygons with a same number of sides are similar if their corresponding angles are equal, and their corresponding sides are in the same ratio.

- Two triangles are similar, if their corresponding angles are equal and corresponding sides are in the same ratio.

- In two triangles, if the corresponding angles are equal and corresponding sides are in the same ratio then the two triangles are similar.

- In two triangles, if the sides of any one of the triangles are proportional to the sides of the other triangle, then their corresponding angles are equal, and the two triangles are similar.

### Theorem 1

To Prove that, If a line is drawn parallel to one side of a triangle and intersects the other two sides, then the other two sides are divided in the same ratio.

ABC is a triangle, side DE || BC and DE intersects AB at D and AC at E.

Join C to D and B to E. Draw EM ⊥ AC and DN ⊥ AB.

We need to prove that AD/DB = AE/EC

Proof:

Area of a triangle = ½ * AD * EM

Similarly,

Ar(BDE) = ½ * DB * EM

Ar(ADE) = ½* AE * DN

Ar(DEC) = ½ * EC * DN

Hence,

Ar(ADE)/Ar(BDE) = ½ * AD * EM / ½ * DB * EM = AD/DB

Similarly,

Ar(ADE)/Ar(DEC) = AE/EC

Triangles DEC and BDE are on the same base, i.e. DE and between same parallels DE and BC.

Hence,

Ar(BDE) = Ar(DEC)

From above equations, we can say that

AD/DB = AE/EC.

Hence Proved

### Theorem 2

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

ABC is a triangle in which DE divides AC and AB in the same ratio. This states that:

AB/DB = AE/EC

To prove: If DE` || BC, then

AB/DB = AE`/E`C

According to the theorem AB/DB = AE/EC

Then Accordingly, E and E` must be coincident.

This proves that DE || BC

Proved

### Theorem 3

In two triangles, If the sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar, also called SSS (side-side-side) criterion.

Two triangles ABC and DEF are drawn in such a way that their corresponding sides are proportional. It means:

AB/DE = AC/DF = BC/EF

To prove: ∠A = ∠C, ∠B = E and ∠C = ∠F

Hence triangle ABC ~ DEF

In Triangle DEF, draw a line PQ so that DP = AB and DQ = AC

Proof:

because the corresponding sides of the two triangles are equal.

This implies;

DP/PE = DQ/QF = PQ/EF

This also means that ∠P = ∠E and ∠Q = ∠F

We had taken, ∠A=∠D, ∠B=∠P and ∠C=∠Q

Hence; ∠A = ∠D, ∠B = ∠E and ∠C = ∠F

Therefore, from AAA criterion;

Triangle ABC ~ DEF.

Proved

### Theorem 4

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

ABC is a triangle which is right angled at B. BD is perpendicular to hypotenuse AC drawn from vertex B.

We have to prove that

Proof:

In triangle ABC and ADB

AB/AC = AD/AB or

Because these are similar triangles.

In triangles ABC and BDC;

BC/AC = CD/BC

When we add 1 and 2, we get;

Proved.