If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form -Maths 9th

If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form -Maths 9th
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According to question the bisectors of the angles APQ, BPQ, CQP and PQD form
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If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form -Maths 9th

Description : If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form -Maths 9th

Last Answer : According to question the bisectors of the angles APQ, BPQ, CQP and PQD form

If two parallel lines are intersected by a transversal, then the bisectors of the interior angles form which one of the following? -Maths 9th

Description : If two parallel lines are intersected by a transversal, then the bisectors of the interior angles form which one of the following? -Maths 9th

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AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Description : AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Last Answer : Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively. To prove AP|| BQ Proof Since, l || m and t is transversal. Therefore, ∠EAB = ∠ABH [alternate interior ... ∠PAB and ∠ABQ are alternate interior angles with two lines AP and BQ and transversal AB. Hence, AP || BQ.

AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Description : AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Last Answer : Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively. To prove AP|| BQ Proof Since, l || m and t is transversal. Therefore, ∠EAB = ∠ABH [alternate interior ... ∠PAB and ∠ABQ are alternate interior angles with two lines AP and BQ and transversal AB. Hence, AP || BQ.

If a transversal intersects two parallel lines, prove that the bisectors of any pair of corresponding angles so formed are parallel. -Maths 9th

Description : If a transversal intersects two parallel lines, prove that the bisectors of any pair of corresponding angles so formed are parallel. -Maths 9th

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Two parallel lines l and m are intersected by a transversal p (see Fig. 8.46). Show that the quadrilateral formed by the bisectors of interior angles is a rectangle. -Maths 9th

Description : Two parallel lines l and m are intersected by a transversal p (see Fig. 8.46). Show that the quadrilateral formed by the bisectors of interior angles is a rectangle. -Maths 9th

Last Answer : Solution :-

In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

Description : In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

Last Answer : Given, In the figure AP|| BQ, AP and BQ are the bisectors of alternate interior angles ∠CAB and ∠ABF. To show l || m Proof Since, AP|| BQ and t is transversal, therefore ∠PAB = ∠ABQ [alternate interior angles] ⇒ 2 ∠PAB = 2 ∠ABQ [multiplying both sides by 2]

In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

Description : In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

Last Answer : Given, In the figure AP|| BQ, AP and BQ are the bisectors of alternate interior angles ∠CAB and ∠ABF. To show l || m Proof Since, AP|| BQ and t is transversal, therefore ∠PAB = ∠ABQ [alternate interior angles] ⇒ 2 ∠PAB = 2 ∠ABQ [multiplying both sides by 2]

In Fig. 9.30, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar(APQ). -Maths 9th

Description : In Fig. 9.30, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar(APQ). -Maths 9th

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

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Last Answer : Given : A parallelogram ABCD such that the bisectors of adjacent angles A and B intersect at P. To prove : ∠APB = 90° Proof : Since ABCD is a | | gm ∴ AD | | BC ⇒ ∠A + ∠B = 180° [sum of consecutive interior ... 90° + ∠APB + ∠2 = 180° [ ∵ ∠1 + ∠2 = 90° from (i)] Hence, ∠APB = 90°

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

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

Last Answer : Given : A parallelogram ABCD such that the bisectors of adjacent angles A and B intersect at P. To prove : ∠APB = 90° Proof : Since ABCD is a | | gm ∴ AD | | BC ⇒ ∠A + ∠B = 180° [sum of consecutive interior ... 90° + ∠APB + ∠2 = 180° [ ∵ ∠1 + ∠2 = 90° from (i)] Hence, ∠APB = 90°

Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. -Maths 9th

Description : Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. -Maths 9th

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Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle. -Maths 9th

Description : Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle. -Maths 9th

Last Answer : Given Let ABCD be a parallelogram and AP, BR, CR, be are the bisectors of ∠A, ∠B, ∠C and ∠D, respectively. To prove Quadrilateral PQRS is a rectangle. Proof Since, ABCD is a parallelogram, then DC ... and ∠PSR = 90° Thus, PQRS is a quadrilateral whose each angle is 90°. Hence, PQRS is a rectangle.

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

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

Last Answer : Given, ABCD is a cyclic quadrilateral. DP and QB are the bisectors of ∠D and ∠B, respectively. To prove PQ is the diameter of a circle. Construction Join QD and QC.

Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. -Maths 9th

Description : Bisectors of the angles B and C of an isosceles triangle with AB = AC intersect each other at O. -Maths 9th

Last Answer : Solution of this question

Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle. -Maths 9th

Description : Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle. -Maths 9th

Last Answer : Given Let ABCD be a parallelogram and AP, BR, CR, be are the bisectors of ∠A, ∠B, ∠C and ∠D, respectively. To prove Quadrilateral PQRS is a rectangle. Proof Since, ABCD is a parallelogram, then DC ... and ∠PSR = 90° Thus, PQRS is a quadrilateral whose each angle is 90°. Hence, PQRS is a rectangle.

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

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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Bisectors of angles A, B and C of a triangle ABC intersects its circumcircle at D, E and F respectively. Prove that angles of triangle DEF are 90° - A/2, 90° - B/2 and 90° - C/2. -Maths 9th

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The bisectors of the angles of a triangle ABC meet BC, CA and AB at X, Y and Z respectively. -Maths 9th

Description : The bisectors of the angles of a triangle ABC meet BC, CA and AB at X, Y and Z respectively. -Maths 9th

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EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

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Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

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Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

Prove that the angle bisectors of a parallelogram form a rectangle. -Maths 9th

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A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent. -Maths 9th

Description : A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent. -Maths 9th

Last Answer : According to question prove that the perpendicular bisectors of AB, BC and CA are concurrent.

A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent. -Maths 9th

Description : A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent. -Maths 9th

Last Answer : According to question prove that the perpendicular bisectors of AB, BC and CA are concurrent.

In Fig.6.29, DE||QR and AP and BP are bisectors of ZEAB and ZRBA respectively. Find ZAPB. -Maths 9th

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Prove that quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. -Maths 9th

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If two lines intersect prove that the vertically opposite angles are equal. -Maths 9th

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Cbqs (case base study ) of chapter 6 Lines and Angles of maths class 9th -Maths 9th

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The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD. -Maths 9th

Description : The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD. -Maths 9th

Last Answer : area of parallelogram=2× area of triangle ABC =2×15=30sq cm theorem on area

ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Last Answer : AB =AC(given) Angle ABC =angle ACB (angle opposite to equal sides) Angle PAC=Angle ABC +angle ACB (Exterior angle property) Angle PAC =2 angle ACB - - - - - - (1) AD BISECTS ANGLE PAC. ANGLE ... AND AC IS TRANSVERSAL BC||AD BA||CD (GIVEN ) THEREFORE ABCD IS A PARALLEGRAM. HENCE PROVED........

ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

Description : ABC is an isosceles triangle in which AB=AC.AD bisects exterior angles PAC and CD parallel AB.Prove that-i)angle DAC=angle BAC ii)∆BCD is a parallelogram -Maths 9th

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Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other. -Maths 9th

Description : Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other. -Maths 9th

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A transversal intersects two parallel lines. -Maths 9th

Description : A transversal intersects two parallel lines. -Maths 9th

Last Answer : Given Two lines AB and CD are parallel and intersected by transversal t at P and 0, respectively. Also, EP and FQ are the bisectors of angles ∠APG and ∠CQP, respectively.

Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other. -Maths 9th

Description : Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other. -Maths 9th

Last Answer : Given Two lines m and n are parallel and another two lines p and q are respectively perpendicular to m and n. i.e., p ⊥ m, p ⊥ n, q ⊥ m, q ⊥ n To prove p||g Proof Since, ... of interior angles on the same side of the transversal is supplementary, then the two lines are parallel. Hence, p||g.

A transversal intersects two parallel lines. -Maths 9th

Description : A transversal intersects two parallel lines. -Maths 9th

Last Answer : Given Two lines AB and CD are parallel and intersected by transversal t at P and 0, respectively. Also, EP and FQ are the bisectors of angles ∠APG and ∠CQP, respectively.

A transversal intersects two lines in such a way that the two interior angle on the same side of transversal are equal.Will the two lines always be parallel? -Maths 9th

Description : A transversal intersects two lines in such a way that the two interior angle on the same side of transversal are equal.Will the two lines always be parallel? -Maths 9th

Last Answer : Solution :- The two lines will not be always parallel as the sum of the two equal angles will not always be 180°. Lines will be parallel when each of the equal angles is equal to 90°.

In Fig.6.5,which of the two lines are parallel? -Maths 9th

Description : In Fig.6.5,which of the two lines are parallel? -Maths 9th

Last Answer : Solution :- l||m, because angles on the same side of the transversal are supplementary, i.e., 128° +52° = 180°. Therefore p is not parallel to q, because 105° + 74° = 179°.

PQ and RS are two equal and parallel line segments.Any points M not lying on PQ or RS is joined to Q and S and lines through P parallel to SM meet at N.Prove that line segments MN and PQ are equal and parallel to each other. -Maths 9th

Description : PQ and RS are two equal and parallel line segments.Any points M not lying on PQ or RS is joined to Q and S and lines through P parallel to SM meet at N.Prove that line segments MN and PQ are equal and parallel to each other. -Maths 9th

Last Answer : hope its clear

l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF

l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF

Does Euclid's fifth postulate imply the existence of parallel lines?Explain. -Maths 9th

Description : Does Euclid's fifth postulate imply the existence of parallel lines?Explain. -Maths 9th

Last Answer : Solution :-

In Fig.6.6, find the value of x for which the lines l and m are parallel. -Maths 9th

Description : In Fig.6.6, find the value of x for which the lines l and m are parallel. -Maths 9th

Last Answer : Solution :-

For what value of x will the lines l and m be parallel to each other ? -Maths 9th

Description : For what value of x will the lines l and m be parallel to each other ? -Maths 9th

Last Answer : Solution :-

l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th

Description : l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th

Last Answer : Given:l∥m∥n l,m and n cut off equal intercepts AB and BC on p So,AB=BC To prove:l,m and n cut off equal intercepts DE and EF on q i.e.,DE=EF Proof:In △ACF, B is the mid-point of ... a triangle, parallel to another side, bisects the third side. Since E is the mid-point of DF DE=EF Hence proved.