Class 10 Maths Theorems Chapter 6

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Chapter 6 Theorems bY RDS EDUCATION



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.
class ten math triangles1-Theorem 1
Construction: ABC is a triangle. DE || BC and DE intersects AB at D and AC at E.
Join B to E and C to D. Draw DN ⊥ AB and EM ⊥ AC.
To prove:
class ten math triangles1-Theorem 1
Proof:
Area of triangle class ten math triangles2-Theorem 1
Similarly,
class ten math triangles3-Theorem 1 
class ten math triangles4-Theorem 1 
class ten math triangles5-Theorem 1
Hence,
class ten math triangles6-Theorem 1
Similarly,
class ten math triangles7-Theorem 1
Triangles BDE and DEC are on the same base, i.e. DE and between same parallels, i.e. DE and BC.
Hence,
class ten math triangles8-Theorem 1
From above equations, it is clear that;
class ten math triangles9-Theorem 1


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.
class ten math triangles1-Theorem 2
Construction: ABC is a triangle in which line DE divides AB and AC in the same ratio. This means:
class ten math triangles10-Theorem 2
To prove: DE || BC
Let us assume that DE is not parallel to BC. Let us draw another line DE' which is parallel to BC.
Proof:
If DE' || BC, then we have;
class ten math triangles11-Theorem 2
According to the theorem;
class ten math triangles12-Theorem 2
Then according to the first theorem; E and E' must be coincident.
This proves: DE || BC


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. This is also called AAA (Angle-Angle-Angle) criterion.
class ten math triangles1-Theorem 3
Construction: Two triangles ABC and DEF are drawn so that their corresponding angles are equal. This means:
∠ A =∠ D, ∠ B = ∠ E and ∠ C = ∠ F
To prove:
class ten math triangles13-Theorem 3
Draw a line PQ in the second triangle so that DP = AB and PQ = AC
Proof:
class ten math triangles14-Theorem 3
Because corresponding sides of these two triangles are equal
This means; ∠ B = ∠ P = ∠ E and PQ || EF
This means;
class ten math triangles15-Theorem 3
Hence;
class ten math triangles16-Theorem 3 
class ten math triangles17-Theorem 3
Hence,
class ten math triangles18-Theorem 3


THEOREM 4:
If in two triangles, 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. This is also called SSS (Side-Side-Side) criterion.
class ten math triangles1-Theorem 3
Construction: Two triangles ABC and DEF are drawn so that their corresponding sides are proportional. This means:
class ten math triangles19-Theorem 4
To prove: ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F
And hence; ▲ABC ~ ▲DEF
In triangle DEF, draw a line PQ so that DP = AB and DQ = AC
Proof:
class ten math triangles20-Theorem 4
Because corresponding sides of these two triangles are equal
This means;
class ten math triangles21-Theorem 4
This also means; ∠ P = ∠ E and ∠ Q = ∠ F
We have taken; ∠ A = ∠ D, ∠ B = ∠ P and ∠ C = ∠ Q
Hence; ∠ A = ∠ D, ∠ B = ∠ E and ∠ C = ∠ F
From AAA criterion;
▲ABC ~ ▲DEF proved

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. This is also called SAS (Side-Angle-Side) criterion.
class ten math triangles1-Theorem 3
Construction: Two triangles ABC and DEF are drawn so that one of the angles of one triangle is equal to one of the angles of another triangle. Moreover, two sides included in that angle of one triangle are proportional to two sides included in that angle of another triangle. This means;
∠ A = ∠ D and
class ten math triangles22-Theorem 4
To prove: ▲ABC ~ ▲DEF
Draw PQ in triangle DEF so that, AB = DP and AC = DF
Proof:
class ten math triangles23-Theorem 5
Because corresponding sides of these two triangles are equal
class ten math triangles24-Theorem 5  given 
∠ A = ∠ D
Hence; class ten math triangles25-Theorem 5  ~ from SSS criterion
Hence;
class ten math triangles26-Theorem 5
Hence; ▲ABC ~ ▲DEF proved

THEOREM 6:
The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
class ten math triangles1-Theorem 5
Construction: Two triangles ABC and PQR are drawn so that, ▲ABC ~ ▲PQR.
To prove:
class ten math triangles27-Theorem 6
Draw AD ⊥ BC and PM ⊥ PR
Proof:
class ten math triangles28-Theorem 6 
class ten math triangles29-Theorem 6
Hence;
class ten math triangles30-Theorem 6
Now, in ▲ABD and ▲PQM;
∠ A = ∠ P, ∠ B = ∠ Q and ∠ D = ∠ M (because ▲ABC ~ ▲PQR)
Hence; ▲ABD ~ ▲PQM
Hence;
class ten math triangles31-Theorem 6
Since, ▲ABC ~ ▲PQR
So,
class ten math triangles32-Theorem 6
Hence;
class ten math triangles33-Theorem 6 class ten math triangles34-Theorem 6
Similarly, following can be proven:
class ten math triangles35-Theorem 6
THEOREM 7:
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.
class ten math triangles1-Theorem 7
Construction: Triangle ABC is drawn which is right-angled at B. From vertex B, perpendicular BD is drawn on hypotenuse AC. To prove:
▲ABC ~  ▲ADB ~ ▲BDC
Proof:
In ▲ABC and ▲ADB;
∠ ABC = ∠ ADB
∠ BAC = ∠ DAB
∠ ACB = ∠ DBA
From AAA criterion; ▲ABC ~ ▲ADB
In ▲ABC and ▲BDC;
∠ ABC = ∠ BDC
∠ BAC = ∠ DBC
∠ ACB = ∠ DBC
From AAA criterion; ▲ABC ~ ▲BDC
Hence; ▲ABC ~ ▲ADB ~ ▲BDC proved.

THEOREM 8:
Pythagoras Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
class ten math triangles1-Theorem 8
Construction: Triangle ABC is drawn which is right angled at B. From vertex B, perpendicular BD is drawn on hypotenuse AC. To prove:
class ten math triangles36-Theorem 6
Proof:
In ▲ABC and ▲ADB;
class ten math triangles37-Theorem 6 
class ten math triangles38-Theorem 6
Because these are similar triangles (as per previous theorem)
In ▲ABC and ▲BDC;
class ten math triangles38-Theorem 6 
class ten math triangles40-Theorem 6
Adding equations (1) and (2), we get;
class ten math triangles41-Theorem 6 
Proved. 

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