Let ABC be a triangle. Let D, E, F be points respectively on segments BC, CA, AB such that AD, BE and CF concur at point K. -Maths 9th

Let ABC be a triangle. Let D, E, F be points respectively on segments BC, CA, AB such that AD, BE and CF concur at point K. -Maths 9th
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Let O be any point inside a triangle ABC. Let L, M and N be the points on AB, BC and CA respectively, -Maths 9th

Description : Let O be any point inside a triangle ABC. Let L, M and N be the points on AB, BC and CA respectively, -Maths 9th

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D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Description : D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Last Answer : Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it. DE = 1 / 2 AC ⇒ DE = AF = CF EF = 1 / 2 AB ⇒ EF = AD = BD DF = 1 ... △DEF ≅ △AFD Thus, △DEF ≅ △CFE ≅ △BDE ≅ △AFD Hence, △ABC is divided into four congruent triangles.

In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Description : In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .

D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Description : D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. -Maths 9th

Last Answer : Since the segment joining the mid points of any two sides of a triangle is half the third side and parallel to it. DE = 1 / 2 AC ⇒ DE = AF = CF EF = 1 / 2 AB ⇒ EF = AD = BD DF = 1 ... △DEF ≅ △AFD Thus, △DEF ≅ △CFE ≅ △BDE ≅ △AFD Hence, △ABC is divided into four congruent triangles.

In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Description : In the fig, D, E and F are, respectively the mid-points of sides BC, CA and AB of an equilateral triangle ABC. -Maths 9th

Last Answer : Since line segment joining the mid-points of two sides of a triangle is half of the third side. Therefore, D and E are mid-points of BC and AC respectively. ⇒ DE = 1 / 2 AB --- (i) E and F are the mid - ... CA ⇒ DE = EF = FD [using (i) , (ii) , (iii) ] Hence, DEF is an equilateral triangle .

D,E and F are the mid-points of the sides BC,CA and AB,respectively of an equilateral triangle ABC.Show that △DEF is also an euilateral triangle -Maths 9th

Description : D,E and F are the mid-points of the sides BC,CA and AB,respectively of an equilateral triangle ABC.Show that △DEF is also an euilateral triangle -Maths 9th

Last Answer : Solution :-

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram (ii) quadrilateral BEFC is a parallelogram (iii) AD || CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Description : In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig. 8.22). Show that (i) quadrilateral ABED is a parallelogram ( ... CF and AD = CF (iv) quadrilateral ACFD is a parallelogram (v) AC = DF (vi) ΔABC ≅ ΔDEF. -Maths 9th

Last Answer : . Solution: (i) AB = DE and AB || DE (Given) Two opposite sides of a quadrilateral are equal and parallel to each other. Thus, quadrilateral ABED is a parallelogram (ii) Again BC = EF and BC || EF ... (Given) BC = EF (Given) AC = DF (Opposite sides of a parallelogram) , ΔABC ≅ ΔDEF [SSS congruency]

Side AC of a right triangle ABC is divided into 8 equal parts. Seven line segments parallel to BC are drawn to AB from the points of division. -Maths 9th

Description : Side AC of a right triangle ABC is divided into 8 equal parts. Seven line segments parallel to BC are drawn to AB from the points of division. -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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D and E are respectively the points on the sides AB and AC of a triangle ABC such that AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE || BC. Then, length of DE (in cm) is (a) 2.5 (b) 3 (c) 5 (d) 6

Description : D and E are respectively the points on the sides AB and AC of a triangle ABC such that AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE || BC. Then, length of DE (in cm) is (a) 2.5 (b) 3 (c) 5 (d) 6

Last Answer : (b) 3

ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th

Description : ABC is an isosceles triangle in which altitude BE and CF are drawn to equal sides AC and AB respectively (Fig. 7.15). Show that these altitudes are equal. -Maths 9th

Last Answer : In △ABE and △ACF, we have ∠BEA=∠CFA (Each 90 0 ) ∠A=∠A (Common angle) AB=AC (Given) ∴△ABE≅△ACF (By SAS congruence criteria) ∴BF=CF [C.P.C.T]

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.

ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.

ABCD is a trapezium in which AB || DC and AD = BC. If P, Q, R and S be respectively the mid-points of BA, BD, CD and CA, then PQRS is a -Maths 9th

Description : ABCD is a trapezium in which AB || DC and AD = BC. If P, Q, R and S be respectively the mid-points of BA, BD, CD and CA, then PQRS is a -Maths 9th

Last Answer : Here is your First of all we will draw a quadrilateral ABCD with AD = BC and join AC, BD, P,Q,R,S are the mid points of AB, AC, CD and BD respectively. In the triangle ABC, P and Q are mid points of AB and AC respectively. All sides are equal so PQRS is a Rhombus.

E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF||AB and EF = 1/2 (AB +CD). -Maths 9th

Description : E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF||AB and EF = 1/2 (AB +CD). -Maths 9th

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Let ABC be a triangle of area 16 cm^2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is -Maths 9th

Description : Let ABC be a triangle of area 16 cm^2 . XY is drawn parallel to BC dividing AB in the ratio 3 : 5. If BY is joined, then the area of triangle BXY is -Maths 9th

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ABC and DBC are two triangles on the same BC such that A and D lie on the opposite sides of BC,AB=AC and DB = DC.Show that AD is the perpendicular bisector of BC. -Maths 9th

Description : ABC and DBC are two triangles on the same BC such that A and D lie on the opposite sides of BC,AB=AC and DB = DC.Show that AD is the perpendicular bisector of BC. -Maths 9th

Last Answer : Solution :-

In triangle ABC, D and E are mid-points of the sides BC and AC respectively. Find the length of DE. Prove that DE = 1/2AB. -Maths 9th

Description : In triangle ABC, D and E are mid-points of the sides BC and AC respectively. Find the length of DE. Prove that DE = 1/2AB. -Maths 9th

Last Answer : First Find the points D and E by midpoint formula. (x₂+x₁/2 , y₂+y₁/2) For DE=1/2AB In ΔsCED and CAB ∠ECD=∠ACB and the ratio of the side containing the angle is same i.e, CD=1/2BC ⇒CD/BC=1/2 EC=1/2AC ⇒EC/AC=1/2 ∴,ΔCED~ΔCAB hence the ratio of their corresponding sides will be equal, DE=1/2AB

ABCD is a trapezium in which side AB is parallel to side DC and E is the mid-point of side AD. If F is a point on side BC such that segment -Maths 9th

Description : ABCD is a trapezium in which side AB is parallel to side DC and E is the mid-point of side AD. If F is a point on side BC such that segment -Maths 9th

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ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Description : ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Last Answer : Solution: (i) In ΔACB, M is the midpoint of AB and MD || BC , D is the midpoint of AC (Converse of mid point theorem) (ii) ∠ACB = ∠ADM (Corresponding angles) also, ∠ACB = 90° , ∠ADM = 90° and MD ⊥ AC (iii ... SAS congruency] AM = CM [CPCT] also, AM = ½ AB (M is midpoint of AB) Hence, CM = MA = ½ AB

ABC is a triangle right-angled at C. A line through the mid-point of hypotenuse AB and parallel to BC intersects AC at D. Show that -Maths 9th

Description : ABC is a triangle right-angled at C. A line through the mid-point of hypotenuse AB and parallel to BC intersects AC at D. Show that -Maths 9th

Last Answer : Solution :-

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Description : 5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively. To show, AF and EC trisect the diagonal BD. Proof, ABCD is a parallelogram , AB || CD also, ... (i), DP = PQ = BQ Hence, the line segments AF and EC trisect the diagonal BD. Hence Proved.

E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. -Maths 9th

Description : E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. -Maths 9th

Last Answer : According to question the mid-points of the non-parallel sides AD and BC of a trapezium ABCD.

E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. -Maths 9th

Description : E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. -Maths 9th

Last Answer : According to question the mid-points of the non-parallel sides AD and BC of a trapezium ABCD.

If a + b + c = 9 and ab + bc + ca = 23, then a3 + b3 + c3 – 3 abc = (a) 108 (b) 207 (c) 669 (d) 729 -Maths 9th

Description : If a + b + c = 9 and ab + bc + ca = 23, then a3 + b3 + c3 – 3 abc = (a) 108 (b) 207 (c) 669 (d) 729 -Maths 9th

Last Answer : a+b+c=9 and a2+b2+c2=35 Using formula, (a+b+c)2=a2+b2+c2+2(ab+bc+ca) 92=35+2(ab+bc+ca) 2(ab+bc+ca)=81−35=46 (ab+bc+ca)=23 using formula, (a3+b3+c3)−3abc=(a2+b2+c2−ab−bc−ca)(a+b+c) a3+b3+c3−3abc=(35−23)×9=9×12=108

In the figure, arcs and drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of teh shaded region. [use π = 3.14] -Maths 10th

Description : In the figure, arcs and drawn by taking vertices A, B and C of an equilateral triangle of side 10 cm to intersect the sides BC, CA and AB at their respective mid-points D, E and F. Find the area of teh shaded region. [use π = 3.14] -Maths 10th

Last Answer : Step-by-step explanation: We have been provided that, Triangle ABC is an Equilateral triangle. Side of triangle is = 10 cm The arcs are drawn from each vertices of the triangle. We get three sectors ... portion is, Remaining area = Area of triangle ABC - Area of all the sectors 39.25cm square

in triangle abc if bd =1/3 bc then prove that 9(ad -Maths 9th

Description : in triangle abc if bd =1/3 bc then prove that 9(ad -Maths 9th

Last Answer : This answer was deleted by our moderators...

in triangle abc if bd =1/3 bc then prove that 9(ad -Maths 9th

Description : in triangle abc if bd =1/3 bc then prove that 9(ad -Maths 9th

Last Answer : This answer was deleted by our moderators...

If in equilateral triangle ABC, AD is perpendicular on BC then Prove that 3ABsquar=4ADsquare -Maths 9th

Description : If in equilateral triangle ABC, AD is perpendicular on BC then Prove that 3ABsquar=4ADsquare -Maths 9th

Last Answer : This answer was deleted by our moderators...

In an equilateral triangle ABC, the side BC is trisected at D. Then AD^2 is equal to -Maths 9th

Description : In an equilateral triangle ABC, the side BC is trisected at D. Then AD^2 is equal to -Maths 9th

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In the given figure, ABC is an equilateral triangle of side length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. -Maths 9th

Description : In the given figure, ABC is an equilateral triangle of side length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. -Maths 9th

Last Answer : answer:

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Description : 4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. To prove, F is the mid-point of BC. Proof, BD intersected EF at G. In ΔBAD, E is the ... point of BD and also GF || AB || DC. Thus, F is the mid point of BC (Converse of mid point theorem)

In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th

Description : In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th

Last Answer : Given, E is mid point of AD Also EB∥DF ⇒ B is mid point of AF [mid--point theorem] so, AF=2AB (1) Since, ABCD is a parallelogram, CD=AB ⇒AF=2CD AD∥BC⇒LB∥AD In ΔFDA ⇒LB∥AD ⇒LDLF​=ABFB​=1 from (1) ⇒LF=LD so, DF=2DL

From a point O in the interior of a DABC if perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively, then which of the -Maths 9th

Description : From a point O in the interior of a DABC if perpendiculars OD, OE and OF are drawn to the sides BC, CA and AB respectively, then which of the -Maths 9th

Last Answer : (i) In Δ O C E ,D C 2 = D E 2 + E C 2 Δ O B D , D B 2 = O D 2 + B D 2 Δ O A F , O A 2 = O F 2 + A F 2 Adding we get O A 2 + O B 2 + O C 2 = O F 2 + O D 2 + O F 2 + E C 2 + B D 2 + A F 2 A F 2 + B D 2 + C E 2 = O A

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

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........

in triangle abc bd =1/3 bd then prove that 9(ad)^2=7(ab)^2 -Maths 9th

Description : in triangle abc bd =1/3 bd then prove that 9(ad)^2=7(ab)^2 -Maths 9th

Last Answer : This answer was deleted by our moderators...

In the above figure ( not to scale ) the sides BA,BC and CA of` Delta ABC` are produced to D,F, and E respectively such that `/_ ACF= 120^(@)` and `/_

Description : In the above figure ( not to scale ) the sides BA,BC and CA of` Delta ABC` are produced to D,F, and E respectively such that `/_ ACF= 120^(@)` and `/_

Last Answer : In the above figure ( not to scale ) the sides BA,BC and CA of` Delta ABC` are produced to D,F, and E ... /_ BAE= 150^(@)`. Then `/_ ABC = ________`.

The median BE and CF of a triangle ABC intersect at G. -Maths 9th

Description : The median BE and CF of a triangle ABC intersect at G. -Maths 9th

Last Answer : According to question the area of ΔGBC = area of the quadrilateral AFGE.

The median BE and CF of a triangle ABC intersect at G. -Maths 9th

Description : The median BE and CF of a triangle ABC intersect at G. -Maths 9th

Last Answer : According to question the area of ΔGBC = area of the quadrilateral AFGE.

If a, b, c are the sides of a triangle and a^2 + b^2 + c^2 = bc + ca + ab, then the triangle is: -Maths 9th

Description : If a, b, c are the sides of a triangle and a^2 + b^2 + c^2 = bc + ca + ab, then the triangle is: -Maths 9th

Last Answer : answer:

In a trapezoid ABCD, side BC is parallel to side AD. Also, the lengths of the sides AB, BC, CD and AD are 8, 2, 8 and 10 units respectively -Maths 9th

Description : In a trapezoid ABCD, side BC is parallel to side AD. Also, the lengths of the sides AB, BC, CD and AD are 8, 2, 8 and 10 units respectively -Maths 9th

Last Answer : answer:

Let a, b, c be positive numbers lying in the interval (0, 1], then a/(1+b+ca)+b/(a+c+ab)+c/(1+a+bc) is -Maths 9th

Description : Let a, b, c be positive numbers lying in the interval (0, 1], then a/(1+b+ca)+b/(a+c+ab)+c/(1+a+bc) is -Maths 9th

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ABCD is a parallelogram AE pependicular to DC CF perpendixular to AD AB =16 m ,AE =8m ,CF =10m ,fimd AD -Maths 9th

Description : ABCD is a parallelogram AE pependicular to DC CF perpendixular to AD AB =16 m ,AE =8m ,CF =10m ,fimd AD -Maths 9th

Last Answer : This answer was deleted by our moderators...

In a parallelogram ABCD, AE is perpendicular to DC and CF is perpendicular to AD. If AB = 10 cm, AE = 6 cm and CF = 8 cm, then find AD. -Maths 9th

Description : In a parallelogram ABCD, AE is perpendicular to DC and CF is perpendicular to AD. If AB = 10 cm, AE = 6 cm and CF = 8 cm, then find AD. -Maths 9th

Last Answer : Given, Parallelogram ABCD pAE = 8cm AB = 16cm CF = 10cm In a parallelogram, we know that opposite sides are equal. Therefore, CD = AB = 16cm To find the value of AD, the base is multiplied with height. Area of parallelogram = b x h 16 x 8 = AD x 10 128 = 10AD AD = 12.8cm

E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. -Maths 9th

Description : E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. -Maths 9th

Last Answer : According to question diagonal AC of a parallelogram ABCD such that AE = CF.

E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. -Maths 9th

Description : E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. -Maths 9th

Last Answer : According to question diagonal AC of a parallelogram ABCD such that AE = CF.

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.