what- AC is the angle bisector of?

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Description : Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side, if intersect they will intersect on the circumcircle of the triangle. -Maths 9th

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Description : A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

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Description : In triangle ABC, angle B =35° , angle C =65° and the bisector of angle BAC meets BC in X. Arrange AX, BX and CX in descending order. -Maths 9th

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Description : A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

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Description : If the bisector of an angle of a triangle bisects the opposite side, prove that the triangle is isosceles. -Maths 9th

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Description : Prove that, the bisector of any two consecutive angles of parallelogram intersect at right angle. -Maths 9th

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Description : Explain Angle Bisector. -Maths 9th

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Description : In the above `Delta ABC` ( not to scale ), OA is the angle bisector of `/_ BAC` . If `OB=OC,/_OAC=40^(@)` and `/_ ABO=20^(@)`. If `/_ OCB=(1)/(2) /_ A

Last Answer : In the above `Delta ABC` ( not to scale ), OA is the angle bisector of `/_ BAC` . If `OB=OC,/_OAC=40^(@) ... OCB=(1)/(2) /_ ACO,` then find `/_ BOC.`

Description : What are the correct order of steps for constructing an angle bisector using only a straightedge and compass?

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Description : What is the concurrent point of the angle bisector?

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Description : 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

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.

Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB. -Maths 9th

Last Answer : Let AB be a chord of a circle having centre at OPQ be the perpendicular bisector of the chord AB, which intersects at M and it always passes through O. To prove arc PXA ≅ arc PYB Construction Join AP and BP. Proof In ... ΔBPM, AM = MB ∠PMA = ∠PMB PM = PM ∴ ΔAPM s ΔBPM ∴PA = PB ⇒arc PXA ≅ arc PYB

Description : 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

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.

Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB. -Maths 9th

Last Answer : Let AB be a chord of a circle having centre at OPQ be the perpendicular bisector of the chord AB, which intersects at M and it always passes through O. To prove arc PXA ≅ arc PYB Construction Join AP and BP. Proof In ... ΔBPM, AM = MB ∠PMA = ∠PMB PM = PM ∴ ΔAPM s ΔBPM ∴PA = PB ⇒arc PXA ≅ arc PYB

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Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, then prove that arc PXA ≅ arc PYB. -Maths 9th

Last Answer : Solution :- Let AB be a chord of a circle having centre at O. Let PQ be the perpendicular bisector of the chord AB intersect it say at M. Perpendicular bisector of the chord passes through the centre of the circle,i. ... = PM (Common) ∴ △APM ≅ △BPM (SAS) PA = PB (CPCT) Hence, arc PXA ≅ arc PYB

Description : If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord. -Maths 9th

Last Answer : Given: Two circles, with centres O and O' intersect at two points A and B. A AB is the common chord of the two circles and OO' is the line segment joining the centres of the two circles. Let OO' intersect ... 0° Thus, AP = BP and ∠APO = ∠BPO = 9 0° Hence, OO' is the perpendicular bisector of AB.

Description : The line x – 4y = 6 is the perpendicular bisector of the segment AB and the co-ordinates of B are (1, 3). Find the co-ordinates of A. -Maths 9th

Last Answer : Co-ordinates of A are \(\bigg(rac{3 imes9+1 imes5}{3+1},rac{3 imes6+1 imes-2}{3+1}\bigg)\) = \(\bigg(rac{32}{4},rac{16}{4}\bigg)\), i.e. (8, 4)Now, \(x\) - 3y + 4 = 0 ⇒ -3y = -\(x\) - 4 ⇒ y = \(rac{x}{3}+rac{4} ... 8) [Using, y - y1 = m (x - x1)]⇒ 3y - 12 = \(x\) - 8 ⇒ 3y - \(x\) = 4.

Description : Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle? -Maths 9th

Last Answer : STEP1: Draw a circle with centre at point O and radius 5 cm. STEP2: Draw its cord AB. STEP3: With centre A as centre and radius more than half of AB, draw two arcs, one on each side ... is perpendicular bisector of AB which is chord of circle, Hence, it passes through the centre of the circle.

Description : Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this line segment. -Maths 9th

Last Answer : Steps of Construction (i) Draw a line segment AB = 5.8 cm. (ii) Taking A as centre and radius more than 1/2AB, draw two arcs, one on either side of AB. (iii) Taking B as centre and ... . (iv) Join CD, intersecting AB at point P. Then, line CPD is the required perpendicular bisector of AB.

Description : what- BD is the perpendicular bisector of?

Last Answer : The head administrator expressed that she will take the antibody once immunization targets are met Statecraft in the countryExecutive Sheik Hasina on Saturday said she has not yet taken the Covid-19 antibody however ... at the Shapla Hall of the Prime Minister's Office (PMO).Likewise Read - PM: D

Description : what- point R is on the perpendicular bisector of QS?

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Description : is this statement true or false If you fold the paper so that A matches up with B and then creases the paper, the line formed by the crease is the perpendicular bisector of AB.?

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Description : is this statement true or falseA perpendicular bisector is the set of points that are equidistant from the endpoints of the bisected segment.?

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Description : Which of these is the definition of a perpendicular bisector?

Last Answer : a line or segment that is perpendicular to the given segment and divides it into two congruent segments

Description : A long conducting wire carrying a current I is bent at `120^(@) `( see figure). The magnetic field B at a point P on the right bisector of bending ang

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Description : What is the perpendicular bisector equation of the line segment whose endpoints are at -4 -10 and 8 -1 on the Cartesian plane?

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Description : If line KL is a bisector of FH at point J which statements is true?

Last Answer : With the limited information provided by the question, the onlystatement which must be true is FJ = JH.

Description : What is the perpendicular bisector equation of a line segment with endpoints of p q and 7p 3q?

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Description : What is the difference between a perpendicular line and a perpendicular bisector?

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Description : What is the perpendicular bisector equation of a line with endpoints of s 2s and 3s 8s?

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Description : What is the perpendicular bisector equation of the line y equals 5x plus 10 spanning the parabola y equals x squared plus 4?

Last Answer : If: y = 5x +10 and y = x^2 +4Then: x^2 +4 = 5x +10Transposing terms: x^2 -5x -6 = 0Factorizing the above: (x-6)(X+1) = 0 meaning x = 6 or x =-1Therefore by substitution endpoints of the line are ... .5 = -1/5(x-2.25) => 5y= -x+114.75Perpendicular bisector equation in its general form: x+5y-114.75= 0

Description : How would the construction be different if you changed the compass setting in the next step of the perpendicular bisector construction?

Last Answer : What is the answer ?

Description : What is the perpendicular bisector equation of a line with endpoints of s 2s and 3s 8s?

Last Answer : Endpoints: (s, 2s) and (3s, 8s)Midpoint: (2s, 5s)Slope of line: 3/1Slope of perpendicular line: -1/3Perpendicular bisector equation: y-5s = -1/3(x-2s) => 3y =-x+17sPerpendicular bisector equation in its general form: x+3y-17s =0

Description : If a point on the perpendicular bisector of a segment, then it is?

Last Answer : Equidistant from the endpoints of the segment.

Description : True or False The intersection of the segment and its bisector is the segment's midpoint.?

Last Answer : TRUE

Description : True or False The intersection of the segment and its bisector is the segment's midpoint.?

Last Answer : TRUE

Description : The point which lies on the perpendicular bisector of the line segment joining the points B(3,5) is: (a) (-3,0) (b) (5,0) (c) (5,-5) (d) (0,0)

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Description : The center of gravity of a triangle lies at the point of (A) Concurrence of the medians (B) Intersection of its altitudes (C) Intersection of bisector of angles (D) Intersection of diagonals

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Description : In a purely cohesive soil, the critical centre lies at the intersection of (A) Perpendicular bisector of slope and the locus of the centre (B) Perpendicular drawn at 1/3rd slope from toe and the ... C) Perpendicular drawn at 2/3rd slope from toe and the locus of the centre (D) Directional angles

Last Answer : Answer: Option D

Description : ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

Last Answer : Given, AC is the common hypotenuse. ∠B = ∠D = 90°. To prove, ∠CAD = ∠CBD Proof: Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi ... D are concyclic. Thus, CD is the chord. ⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)

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........

Description : ABC and ADC are two right triangles with common hypotenuse AC. Prove that angle CAD = angle CAB -Maths 9th

Last Answer : Given, AC is the common hypotenuse. ∠B = ∠D = 90°. To prove, ∠CAD = ∠CBD Proof: Since, ∠ABC and ∠ADC are 90°. These angles are in the semi circle. Thus, both the triangles are lying in the semi ... D are concyclic. Thus, CD is the chord. ⇒ ∠CAD = ∠CBD (Angles in the same segment of the circle)