Proofs of Some Important Triangle Theorems:

Proofs of Some Important Triangle Theorems:

Theorem 1:

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, DE is a line parallel to BC and intersecting AB at D and AC at E, i.e. DE || BC.

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

Proof:

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

Proof:

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

Since 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.

Hypotenuse2 = Base2 + Perpendicular2

Proof:

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

To prove: AC2 = AB2 + BC2

In triangle ABC and ADB,

AB/AC = AD/AB

AB2 = AC × AD ……………(1) [Since these are similar triangles.]

In triangles ABC and BDC;

BC/AC = CD/BC

BC2 = AC × CD ………….(2)

When we add 1 and 2, we get;

AB2 + BC2 = AD × AC + CD × AC

AB2 + BC2 = AC (AD + CD)

AB2 + BC2 = AC2

Hence, Proved.