Why parallel line can not intersect -Maths 9th

Why parallel line can not intersect -Maths 9th
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Its a simple question which have a simple answer.. The answer is that :- When you draw a straight line on a paper you will find that the line is straight and when you measure it by you protector or D you will find that the line is 180° which means this line goes straight and long... And when you draw a line at the same angle and between below to it by giving space from the previous line you draw. The second point was that:- As you know about perpendicular bisector.        It is also done by it. When you draw parallel line then do perpendicular bisector (roughly) on it you will find 90° on both side of the line so that means      90°+90°=180° which means it will never interest each other at any condition...
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Related questions

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Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E.

Why parallel line can not intersect -Maths 9th

Description : Why parallel line can not intersect -Maths 9th

Last Answer : Its a simple question which have a simple answer.. The answer is that :- When you draw a straight line on a paper you will find that the line is straight and when you measure it by you protector or ... so that means 90°+90°=180° which means it will never interest each other at any condition...

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Last Answer : Solution :- Construction: Draw two circles having centres O and O' intersecting at points A and B. Draw a parallel line PQ to OO' ... iii) Again, OO' = MN [As OO' NM is a rectangle] ...(iv) ⇒ 2OO' = PQ Hence proved.

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Last Answer : In the given figure: Area of triangle ADC = Area of triangle BCD (Triangles on the same and parallel) Now subtract the area of triangle DOC from both of them so... (Area of triangle ADC - Area of ... => Area of triangle AOD = Area of triangle BOC Hence the area of triangle BOC is 30 cm square.

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Last Answer : ∴ ∠CAO = ∠ODB = x [angles in same segment ] ---- (i) Now, in right angled ΔDOB , ∠ODB + ∠DOB + ∠OBD = 180° ⇒ x + 90° + y =180° (using equation i) ⇒ x + y = 90°

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Description : a squar ABCD in which AC =BE when BC produced .A is joined to E prove that FG=GE when AE intersect BD at F and CD at G -Maths 9th

Last Answer : Please give the figure to get your answer, as it is necessary to have figure to answer the question related to geometry.

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Description : The diagonals AC and BD of parallelogram ABCD intersect at the point O. -Maths 9th

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Description : In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles. -Maths 9th

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ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that : -Maths 9th

Description : ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove that : -Maths 9th

Last Answer : (i) Since diagonals of a parallelogram bisect each other. ∴ O is the mid - point AC as well as BD. In △ADC, OD is a median. ∴ ar(△ADO) = ar(△CDO) [∵ A median of a triangle divide it into two triangles of equal ... and (i) , we have ar(△AOB) - ar(△AOP) = ar(△BOC) - ar(△COP) ⇒ ar(△ABP) = (△CBP)

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. -Maths 9th

Description : Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. -Maths 9th

Last Answer : Draw AM ⟂ BD and CL ⟂ BD. Now, ar(△APB) × ar(△CPD) = {1/2 PB × AM} × {1/2 DP × CL} = {1/2 PB × CL} × {1/2 DP × AM} ar(△BPC) × ar(△APD) Hence, ar(△APB) × ar(△CPD) = ar(△APD) × ar(△BPC)

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Last Answer : Solution of this question

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Description : If two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord. -Maths 9th

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Description : If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle. -Maths 9th

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Description : Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD. -Maths 9th

Last Answer : According to question prove that PB = PD.