Proof of Theorems: Triangles | Mathematics (Maths) Class 10 PDF Download

Theorem 1: Basic Proportionality Theorem (Thales' Theorem)

Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides, then it divides those sides in the same ratio.

Given: In △ABC, DE is drawn parallel to BC such that it intersects AB at D and AC at E, i.e., DE || BC.

To Prove:

ADDB = AEEC

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

Proof:

Area of △ADE = 12 × AD × EM

Area of △BDE = 12 × DB × EM

Area of △ADE = 12 × AE × DN

Area of △DEC = 12 × EC × DN

Therefore, Ar(ADE)Ar(BDE) = ADDB

Similarly, Ar(ADE)Ar(DEC) = AEEC

Triangles DEC and BDE are on the same base (DE) and between the same parallels (DE || BC).

Hence, Ar(BDE) = Ar(DEC).

From the above equations, we conclude:

ADDB = AEEC

Conclusion: The theorem is proved.

Theorem 2: Converse of Basic Proportionality Theorem

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

Given: In △ABC, a line DE divides AB and AC such that:

ADDB = AEEC

To Prove: DE || BC

Construction: Assume a new line DE' parallel to BC, intersecting AC at E'.

Proof:

By Thales' Theorem, since DE' || BC:

ADDB = AE'E'C

From the given condition:

ADDB = AEEC

Since E and E' satisfy the same ratio, they must coincide.

Thus, DE || BC.

Conclusion: The theorem is proved.

Theorem 3: SSS Similarity Criterion

Statement: In two triangles, if the sides of one triangle are proportional to the sides of another triangle, then their corresponding angles are equal, and hence the triangles are similar.

Given: Two triangles △ABC and △DEF such that:

ABDE = ACDF = BCEF

To Prove: △ABC ~ △DEF

Proof:

Since it is given that:

ABDE = ACDF = BCEF

The corresponding sides of △ABC and △DEF are in the same ratio.

We now place △ABC and △DEF such that:

• AB corresponds to DE
• AC corresponds to DF
• BC corresponds to EF

By construction, when the sides of two triangles are proportional, their corresponding **angles must be equal**. Thus, we get:

∠A = ∠D, ∠B = ∠E, ∠C = ∠F

Since all three pairs of corresponding angles are equal, by the **AAA (Angle-Angle-Angle) Similarity Criterion**, we conclude that:

△ABC ~ △DEF

Conclusion: The theorem is proved.

Theorem 4: Pythagoras Theorem

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

Given: In △ABC, ∠B = 90°, AC is the hypotenuse.

To Prove:

AC² = AB² + BC²

Construction: Draw BD ⊥ AC from B to AC.

Proof:

In △ABC and △ADB:

ABAC = ADAB

Multiplying both sides by AC:

AB² = AC × AD

Similarly, in △ABC and △BDC:

BCAC = CDBC

Multiplying both sides by AC:

BC² = AC × CD

Adding both results:

AB² + BC² = AC × AD + AC × CD

Since AD + CD = AC, we get:

AB² + BC² = AC²

Conclusion: The theorem is proved.