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TOPICS
Unit-6(Theorems)

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.

Solution :-
```
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)
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

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 :
```

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.

Solution :-
```
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.
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.

Solution :-
```
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/DF

Note : 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.

Solution :-
```
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.

Solution :-
```
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 DEF

Note :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.

Solution :-
```
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 90o)
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.

Solution :-
```
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.

Solution :-
```
Given that : We are given a right triangle ABC right angled at B.
Prove that : AC2 = AB2 + BC2
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 = AB2 .......(i)
Also, Δ BDC ~ Δ ABC (Theorem 7)
So, CD/BC = BC/AC
or, CD . AC = BC2 .......(ii)
Adding (1) and (2), AD . AC + CD . AC = AB2 + BC2
or, AC (AD + CD) = AB2 + BC2
or, AC . AC = AB2 + BC2
or, AC2 = AB2 + BC2
```

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.

Solution :-
```
Given that : We are given a triangle ABC in which AC2 = AB2 + BC2.
Prove that : ∠ B = 90o.
Construction : We construct a Δ PQR right angled at Q such that PQ = AB and QR = BC
Proof : Now, from Δ PQR, we have :
PR2 = PQ2 + QR2 (Pythagoras Theorem, as ∠ Q = 90o)
or, PR2 = AB2 + BC2 (By construction) ....... (i)
But AC2 = AB2 + BC2 (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 = 90o (By Construction)
So, ∠ B = 90o
```
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