Question

QUESTION:-

Prove all the theorem of quadrilateral of NCERT of class 9

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– Non-copied Answer
– Well-explained Answer​​

Answer 1

$\mathtt{\underline{\large{Theorem \: 1:}}}$

The sum of all the four angles of a quadrilateral is 360°

Construction: Join BD

proof

since the sum of the angles of a triangle is 180°. We have

$\sf{\angle A+ \angle 1+\angle 2=180°(sum \: of \angle \: 's \: in \triangle)} \\ \sf{\angle 3 + \angle C + \angle 4 = 180°(sum \: of \angle \: 's \: in \triangle)}$

On adding them we get

$\sf\angle A+ \angle \: C+(\angle1+\angle3)+(\angle2+\angle 4) =360°</p><p>$

$\sf{\angle A+ \angle \: C+\angle B+ \angle D= 360°}$

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$\mathtt{\underline{\large{Theorem 2:}}}$

• each diagnoal divides the parallelogram into 2 congruent Triangle
• opposite sides are equal
• opposite sides are equal

Construction: Join A and C

Proof :

(i) In ∆ ABC and ∆ CDA

• angle 1= angle 2[alternative int. angle]

• angle 3 = angle 4[alternative int. angle]

• AC= CA[ common]

$\sf{ \therefore \: ∆ABD ≅ ∆CBA \: by \: AAS \: criteria.}$

(ii)similarly ∆ABC ≅ ∆CDA

therefore AB= CD & BC= AD ( c.p.c.t)

(iii ) ∆ABC≅ ∆CDA

$\sf{\therefore \: angle \: B = angle \: D \: (c.p.c.t)}$

Also , angle 1 = angle 2 & angle 3 = angle 4

$\sf{\therefore \angle 1 + \angle 4 = \angle 3 +\angle 2 \implies \angle A = \angle C}$

$\sf{Hence \: \angle \: B = \angle\: D \: and \: \angle A = \angle C}$

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$\mathtt{\underline{\large{Theorem \: 3:}}}$

Prove that each angle of a rectangle is a right angle

Given : In rectangle ABCD angle A is 90°

Proof :

ABCD is a rectangle.

$\sf{\implies ABCD \: is \: ||gm} \\ \\ \sf{\implies Ab || BC \: and \: AB \: is \: a \: transversal.}$

• angle A+ angle B = 180° [ co-interior angles ]

• angle B = 180° - angle A

• angle B= 180°- 90° =90°

angle A= angle C = 90° & angle B= angle D= 90°

Thus each angle of rectangle is right angled.

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$\mathtt{\underline{\large{Theorem \: 4:}}}$

(Midpoint Theorem) The line segment joining the midpoint of any two sides is parallel to the third side and equal to its half.

construction: Draw CF|| BA meeting DE produced in F

In ∆ AED and CEF

• angle AED= angle CEF

• AE=CE

• angle DAE=angle FCE

∆ AED≅∆ CEF BY ASA criteria

ED= CF and DE= EF by c.p.c.t

But, AD = BD [ since D is midpoint of AB]

and BD || CF [by construction]

↝BCFD is a ||gm.

↝DF||BC and DF = BC

↝DE || BC and DE = 1/2 DF = 1/2 BC [since DE = EF]

Hence DE || BC and DE = 1/2 BC

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Answer 2

Answer:

Theorems for Class 9 Maths Chapter 8 Quadrilateral

1st Theorem- The diagonal divides the parallelogram into two congruent triangles.

2nd Theorem- The opposite side of a parallelogram are equal.

3rd Theorem- The quadrilaterals in which each pair of opposite sides are equal are called parallelogram.

4th Theorem- The opposite angles of a parallelogram are equal.

5th Theorem- If each pair of opposite angles are equal then it is a parallelogram.

6th Theorem- The diagonals bisect each other in a parallelogram.

7th Theorem- If the diagonals bisect each other in a quadrilateral, then it’s a parallelogram.

8th Theorem- The opposite sides of a parallelogram are parallel and equal.

9th Theorem- The line that joins mid-points of a triangle’s two side is parallel to the third side.