TOPICS

Unit-6(Theorems)

Triangles

**Theorem-1 :-** If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Given that : A Triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively. Prove that : AD/DB = AE/EC Construction : Let us join BE and CD and then draw DM ⊥ AC and EN ⊥ AB. Proof : Now, area of Δ ADE = 1/2 x AD x EN = (1/2 x Base x Height) So, ar(ADE) = 1/2 x AD x EN Similarly, ar(BDE) = 1/2 x BD x EN ar(ADE) = 1/2 x AE x DM and ar(DEC) = 1/2 x EC x DM Therefore, ar(ADE)/ar(BDE) = (1/2 x AD x EN)/(1/2 x BD x EN) = AD/BD .....(i) and ar(ADE)/ar(DEC) = (1/2 x AE x DM)/(1/2 x EC x DM) = AE/CE .....(ii) Note that Δ BDE and DEC are on the same base DE and between the same parallels BC and DE. So, ar(BDE) = ar(DEC) .....(iii) Therefore, from (i), (ii) and (iii), we have : AD/BD = AE/EC

**Theorem-2 :-** If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Proof : Taking a line DE such that AD/DB = AE/EC and assuming that DE is not parallel to BC. If DE is not parallel to BC, draw a line DE' parallel to BC. So, AD/DB = AE'/E'C Therefore, AE/EC = AE'/E'C Adding 1 to both sides of above, you can see that E and E′ must coincide.

**Theorem-3 :-** If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.

Given that : We are taking two triangles ABC and DEF such that ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F Prove that : AB/DE = BC/EF and also AB/DE = BC/EF = AC/DF Construction : Cut DP = AB and DQ = AC and join PQ. Proof : In Δ ABC and Δ DPQ, we get Δ ABC ≅ Δ DPQ (By congruency rule) This gives ∠ B = ∠ P = ∠ E and PQ || EF (By CPCT Rule) Therefore, DP/PE = DQ/QF i.e., AB/DE = AC/DF SImilarly, AB/DE = BC/EF and so AB/DE = BC/EF = AC/DFNote :This criterion is referred to as the AAA (Angle–Angle–Angle) criterion of similarity of two triangles.

**Theorem-4 :-** If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similiar.

Given that : We are taking two triangles ABC and DEF such that AB/DE = BC/EF = AC/DF Prove that : ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F Construction : Cut DP = AB and DQ = AC and join PQ. Proof : It can be seen that DP/PE = DQ/QF and PQ || EF So, ∠ P = ∠ E and ∠ Q = ∠ F Therefore, DP/DE = DQ/DF = BC/EF So, DP/DE = DQ/DF = BC/EF So, BC = PQ We get, Δ ABC ≅ Δ DPQ (By congruency rule) So, ∠ A = ∠ D, ∠ B = ∠ E and∠ C = ∠ F (By CPCT Rule)Note :This criterion is referred to as the SSS (Side–Side–Side) similarity criterion for two triangles.

**Theorem-5 :-** If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

Given that : We are taking two triangles ABC and DEF such that AB/DE = AC/DF and ∠ A = ∠ D Prove that : ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F Construction : Cut DP = AB, DQ = AC and join PQ. Proof : Now, PQ || EF and Δ ABC ≅ Δ DPQ (By Congruency Rule) So, ∠ A = ∠ D, ∠ B = ∠ P and ∠ C = ∠ Q (By CPCT Rule) Therefore, Δ ABC ~ Δ DEFNote :This criterion is referred to as the SAS (Side–Angle–Side) similarity criterion for two triangles.

**Theorem-6 :-** The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Given that : We are given two triangles ABC and PQR such that Δ ABC ~ Δ PQR Prove that : ar(ABC)/ar(PQR) = (AB/PQ)^{2}= (BC/QR)^{2}= (AC/PR)^{2}Construction : We draw altitudes AM and PN of the triangles. Proof : Now, ar(ABC) = 1/2 x BC x AM and ar(PQR) = 1/2 x QR x PN So, ar(ABC)/ar(PQR) = (1/2 x BC x AM)/(1/2 x QR x PN) = (BC x AM)/(QR x PN) ......(i) Now, in Δ ABM and Δ PQN, ∠ B = ∠ Q (As Δ ABC ~ Δ PQR) and ∠ M = ∠ N (Each is of 90^{o}) So, Δ ABM ~ Δ PQN (By AA similarity criterion) Therefore, AM/PN = AB/PQ ......(ii) Also, Δ ABC ~ Δ PQR (Given) So, AB/PQ = BC/QR = CA/RP ......(iii) Therefore, ar(ABC)/ar(PQR) = AB/PQ x AM/PN [from (i) & (iii)] = AB/PQ x AB/PQ [from (ii)] = (AB/PQ)^{2}Now using (iii), we get ar(ABC)/ar(PQR) = (AB/PQ)^{2}= (BC/QR)^{2}= (CA/RP)^{2}

**Theorem-7 :-** In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

Proof : Now, let us take a right triangle ABC, right angled at B. Let BD be the perpendicular to the hypotenuse AC. In Δ ADB and Δ ABC ∠ A = ∠ A and ∠ ADB = ∠ ABC (Why?) So, Δ ADB ~ Δ ABC (AA similarity criterion) .....(i) Similarly, Δ BDC ~ Δ ABC (AA similarity criterion) .....(ii) So, from (i) and (ii), triangles on both sides of the perpendicular BD are similar to the whole triangle ABC. Also, since Δ ADB ~ Δ ABC and Δ BDC ~ Δ ABC So, Δ ADB ~ Δ BDC

**Theorem-8 :-** In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given that : We are given a right triangle ABC right angled at B. Prove that : AC^{2}= AB^{2}+ BC^{2}Construction : Let us draw BD ⊥ AC. Proof : Now, Δ ADB ~ Δ ABC (Theorem 7) So, AD/AB = AB/AC (Sides are proportional) or, AD . AC = AB^{2}.......(i) Also, Δ BDC ~ Δ ABC (Theorem 7) So, CD/BC = BC/AC or, CD . AC = BC^{2}.......(ii) Adding (1) and (2), AD . AC + CD . AC = AB^{2}+ BC^{2}or, AC (AD + CD) = AB^{2}+ BC^{2}or, AC . AC = AB^{2}+ BC^{2}or, AC^{2}= AB^{2}+ BC^{2}

**Theorem-9 :-** In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Given that : We are given a triangle ABC in which AC^{2}= AB^{2}+ BC^{2}. Prove that : ∠ B = 90^{o}. Construction : We construct a Δ PQR right angled at Q such that PQ = AB and QR = BC Proof : Now, from Δ PQR, we have : PR^{2}= PQ^{2}+ QR^{2}(Pythagoras Theorem, as ∠ Q = 90^{o}) or, PR^{2}= AB^{2}+ BC^{2}(By construction) ....... (i) But AC^{2}= AB^{2}+ BC^{2}(Given) ........(ii) So, AC = PR [From (i) and (ii)] .........(iii) Now, in Δ ABC and Δ PQR, AB = PQ (By construction) BC = QR (By construction) AC = PR [Proved in (iii) above] So, Δ ABC ≅ Δ PQR (SSS Congruencey Rule) Therefore, ∠ B = ∠ Q (CPCT) But ∠ Q = 90^{o}(By Construction) So, ∠ B = 90^{o}

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