**Theorems with logical proofs**

**Theorem 1. :**If a ray stands on a line, then the sum of the adjacent angles so formed is 180°.

**Given:**The ray

*PQ*stands on the line

*XY*.

**To Prove:**m∠

*QPX*+ m∠

*YPQ*= 180°.

**Construction:**Draw

*PE*perpendicular to

*XY*.

**Proof:**m∠

*QPX*= m∠

*QPE*+ m∠

*EPX*

*=*m∠

*QPE*+ 90°

**(1)**

m∠

*YPQ*= m∠*YPE*− m∠*QPE**=*90° − m∠

*QPE*

**(2)**

**(1) + (2)**⇒ m∠

*QPX +*m∠

*YPQ =*(m∠

*QPE +*90°) + (90° − m ∠

*QPE*) = 180°.

Thus the theorem is proved.

**Theorem 2:**If two lines intersect, then the vertically opposite angles are of equal measure.

**Given:**Two lines

*AB*and

*CD*intersect at the point

*O*

**To prove:**m∠

*AOC*= m∠

*BOD*, m∠

*BOC*= m∠

*AOD.*

**Proof:**The ray

*OB*stands on the line

*CD*.

∴ m∠

*BOD +*m∠

*BOC*= 180°

**(1)**

The ray

*OC*stands on the line*AB*.∴ m∠

*BOC +*m∠*AOC =*180°**(2)**From

**(1)**and**(2)**,m∠

*BOD*+ m∠*BOC*= m∠*BOC*+ m∠*AOC*∴ m∠

*BOD*= m∠*AOC*.Since the ray

*OA*stands on the line*CD*,m∠

*AOC*+ m∠*AOD*= 180°**(3)**From

**(2)**and**(3)**, we getm∠

*BOC*+ m∠*AOC*= m∠*AOC*+ m∠*AOD*∴ m∠

*BOC*= m∠*AOD*.Hence the theorem is proved.

**Theorem 3:**The sum of the three angles of a triangle is 180°.

**Given:**

*ABC*is a triangle (see Figure 6.54).

**To prove:**∠

*A*+ ∠

*B*+ ∠

*C*= 180°.

**Construction:**Through the vertex

*A*, draw the line

*XY*parallel to the side

*BC.*

**Proof:**

*XY*||

*BC*∴ m∠

*XAB*= m∠

*ABC*(alternate angles).

= ∠

*B.***(1)**Next,

*AC*is a transversal to the parallel lines*XY*and*BC*.∴ m∠

*YAC*= m∠*ACB*(alternate angles) = ∠*C.***(2)**We also have m∠

*BAC*= m∠*A.***(3)****(1) + (2) + (3)**⇒ m∠

*XAB*+ m∠

*YAC*+ m∠

*BAC*= m∠

*B*+ m∠

*C*+ m∠

*A*

⇒ (m∠

*XAB*+ m∠*BAC*) + m∠*CAY*= m∠*A*+ m∠*B*+ m∠*C*⇒ m∠

*XAC*+ m∠*CAY*= m∠*A*+ m∠*B*+ m∠*C*⇒ 180° = m∠

*A*+ m∠*B*+ m∠*C.*Hence the theorem is proved.

**Theorem4. :**The angles opposite to equal sides of a triangle are equal.

**Given:**

*ABC*is a triangle where

*AB*=

*AC*).

**To prove:**∠

*B*= ∠

*C*.

**Construction:**Mark the mid point of

*BC*as

*M*and join

*AM.*

**Proof:**In the triangles

*AMB*and

*AMC*

(i)

*BM*=*CM*(ii)*AB*=*AC*(iii)*AM*is common.∴ By the SSS criterion, Δ

*AMB*≡ Δ*AMC*.∴ Corresponding angles are equal. In particular, ∠

*B*= ∠*C*.Hence the theorem is proved.

**The side opposite to the larger of two angles in a triangle is longer than the side opposite to the smaller angle.**

Theorem 5. :

Theorem 5. :

**Given:**

*ABC*is a triangle, where ∠

*B*is larger than ∠

*C,*that is m∠

*B*> m∠

*C*.

**To prove:**The length of the side

*AC*is longer than the length of the side

*AB.*

i.e.,

*AC*>*AB*(see Figure 6.56).**Proof:**The lengths of

*AB*and

*AC*are positive numbers. So three cases arise

(i)

*AC*<*AB*(ii)

*AC*=*AB*(iii)

*AC*>*AB***Case (i)**Suppose that

*AC*<

*AB*. Then the side

*AB*has longer length than the side

*AC*. So the angle ∠

*C*which is opposite to

*AB*is larger measure than that of ∠

*B*which is opposite to the shorter side

*AC*. That is, m∠

*C*> m∠

*B*. This contradicts the given fact that m∠

*B*> m∠

*C*. Hence the assumption that

*AC*<

*AB*is wrong.

∴

*AC*<*AB.***Case (ii)**Suppose that

*AC*=

*AB*. Then the two sides

*AB*and

*AC*are equal. So the angles opposite to these sides are equal. That is ∠

*B*= ∠

*C*. This is again a contradiction to the given fact that ∠

*B*> ∠

*C*. Hence

*AC*=

*AB*is impossible. Now Case (iii) remains alone to be true.

Hence the theorem is proved.

**Theorem 6. :**A parallelogram is a rhombus if its diagonals are perpendicular.

**Given:**

*ABCD*is a parallelogram where the diagonals

*AC*and

*BD*are perpendicular.

**To prove:**

*ABCD*is a rhombus.

**Construction:**Draw the diagonals

*AC*and

*BD*. Let

*M*be the point of intersection of

*AC*and

*BD*(see

**Proof:**In triangles

*AMB*and

*BMC*,

(i) ∠

*AMB*= ∠*BMC*= 90°(ii)

*AM*=*MC*(iii)

*BM*is common.∴ By SSA criterion, Δ

*AMB*≡ Δ*BMC*.∴Corresponding sides are equal.

In particular,

*AB*=*BC*.Since

*ABCD*is a parallelogram,*AB*=*CD*,*BC*=*AD*.∴

*AB*=*BC*=*CD*=*AD*.Hence

*ABCD*is a rhombus. The theorem is proved.**Theorem 7:**Prove that the bisector of the vertex angle of an isosceles triangle is a median to the base.

**Solution:**Let

*ABC*be an isosceles triangle where

*AB*=

*AC.*Let

*AD*be the bisector of the vertex angle ∠

*A.*We have to prove that

*AD*is the median of the base

*BC.*That is, we have to prove that

*D*is the mid point of

*BC*. In the triangles

*ADB*and

*ADC*,

we have

*AB*=*AC*, m∠*BAD*= m∠*DAC AD*is an angle (bisector),*AD*is common.∴ By SAS criterion, ∠

*ABD*≡ Δ*ACD*.∴ The corresponding sides are equal.

∴

*BD*=*DC*.i.e.,

*D*is the mid point of*BC*.**Theorem 8**Prove that the sum of the four angles of a quadrilateral is 360°.

**Solution:**Let

*ABCD*be the given quadrilateral. We have to prove that ∠

*A*+ ∠

*B*+ ∠

*C*+ ∠

*D*= 360°. Draw the diagonal

*AC*. From the triangles

*ACD*and

*ABC*, we get

∠

*DAC*+ ∠*D*+ ∠*ACD*= 180°**(1)**∠

*CAB*+ ∠*B*+ ∠*ACB*= 180°**(2)****(1)**+

**(2)**⇒ ∠

*DAC*+ ∠

*D*+ ∠

*ACD*+ ∠

*CAB*

+ ∠

*B*+ ∠*ACB*= 360°⇒ (∠

*DAC*+ ∠*CAB*) + ∠*B*+ (∠*ACD*+ ∠*ACB*)+ ∠*D*= 360°⇒ ∠

*A*+ ∠*B*+ ∠*C*+ ∠*D*= 360°.**Theorem**

**8:**In a rhombus, prove that the diagonals bisect each other at right angles.

**Solution:**Let

*ABCD*be a rhombus, Draw the diagonals

*AC*and

*BD*. Let them meet at

*O*. We have to prove that

*O*is the mid point of both

*AC*and

*BD*and that

*AC*is perpendicular (⊥) to

*BD*.

Since a rhombus is a parallelogram, the diagonals

*AC*and

*BD*bisect each other.

∴

*OA*=*OC*,*OB*=*OD.*In triangles

*AOB*and*BOC*, we have(i)

*AB*=*BC*(ii)

*OB*is common(iii)

*OA*=*OC*∴ Δ

*AOB*= Δ*BOC*, by SSS criterion.∴ ∠

*AOB*= ∠*BOC.*Similarly, we can get ∠

*BOC*= ∠*COD*, ∠*COD*= ∠*DOA*.∴ ∠

*AOB*= ∠*BOC*= ∠*COD*= ∠*DOA*=*x*(say)But ∠

*AOB*+ ∠*BOC*+ ∠*COD*+ ∠*DOA*= 360°∴

*x*+*x*+*x*+*x*= 360°∴ 4

*x*= 360° or*x*= 4360° = 90°.∴ The diagonals bisect each other at right angles.

**Theorem**

**9:**Prove that a diagonal of a rhombus bisects each vertex angles through which it passes.

**Solution:**Let

*ABCD*be the given rhombus. Draw the diagonals

*AC*and

*BD.*Since

*AB*||

*CD*and

*AC*is a transversal to

*AB*and

*CD.*We get

∠

*BAC*= ∠*ACD*(alternate angles are equal)**(1)**But

*AD*=*CD*(since*ABCD*is a rhombus)∴Δ

*ADC*is isosceles.∴∠

*ACD*= ∠*DAC*(angles opposite to the equal sides are equal)

**(2)**From

**(1)**and**(2)**, we get∠

*BAC*= ∠*DAC*i.e.,

*AC*bisects the angle ∠*A.*Similarly we can prove that

*AC*bisects ∠*C*,*BD*bisects ∠*B*and*BD*bisects ∠*D*.
Your blog contains all the article which are really important in students studies & also it covers all the detailed points & also they are very easy to understand & very helpful

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ReplyDeletei asked for circle not for triangles and quadrilaterals theorem.

ReplyDeleteFor chapter circle not for triangles...

ReplyDeleteReally useless every one knows all these

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ReplyDeleteI asked for area of parallelogram and triangles not of quadrilateral

ReplyDeleteUselesssssss......'......................

Area og parallelogram = base × height

DeleteArea of triangle = base×height(if triangle is equilateral or isoceles)

Area of right anle triangle= 1÷2×base×height(if trianle is right angle)

These theorems are not of ant use in circle chaptet.This does not contain NCERT book theorem.You made ur own theorems for circle chapter.Unsatisfie,very dissatisfying.Disappoined!:( :(

ReplyDeleteBakwass theorems

ReplyDelete