In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC -Maths 9th

In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC -Maths 9th
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Given, ABCD is a parallelogram. BE = AB To show, ED bisects BC Proof: AB = BE (Given) AB = CD (Opposite sides of ||gm) ∴ BE = CD Let DE intersect BC at F. Now, In ΔCDO and ΔBEO, ∠DCO = ∠EBO (AE || CD) ∠DOC = ∠EOB (Vertically opposite angles) BE = CD (Proved) ΔCDO ≅ ΔBEO by AAS congruence condition. Thus, BF = FC (by CPCT) Therefore, ED bisects BC. Proved
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In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC -Maths 9th

Description : In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC -Maths 9th

Last Answer : Given, ABCD is a parallelogram. BE = AB To show, ED bisects BC Proof: AB = BE (Given) AB = CD (Opposite sides of ||gm) ∴ BE = CD Let DE intersect BC at F. Now, In ΔCDO and ΔBEO, ∠DCO = ... CD (Proved) ΔCDO ≅ ΔBEO by AAS congruence condition. Thus, BF = FC (by CPCT) Therefore, ED bisects BC. Proved

ABCD is a parallelogram. AB is produced to E such that BE = AB. Prove that ED bisects BC. -Maths 9th

Description : ABCD is a parallelogram. AB is produced to E such that BE = AB. Prove that ED bisects BC. -Maths 9th

Last Answer : answer:

ABCD is parallelogram . AB is produced to E so that BE = AB. Provethat ED bisects BC -Maths 9th

Description : ABCD is parallelogram . AB is produced to E so that BE = AB. Provethat ED bisects BC -Maths 9th

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

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

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........

Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus -Maths 9th

Description : Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus -Maths 9th

Last Answer : (i) Here, ABCD is a parallelogram and diagonal AC bisects ∠A. ∴ ∠DAC=∠BAC ---- ( 1 ) Now, AB∥DC and AC as traversal, ∴ ∠BAC=∠DCA [ Alternate angles ] --- ( 2 ) AD∥BC and AAC as traversal, ∴ ∠DAC= ... ---- ( 2 ) From ( 1 ) and ( 2 ), ⇒ AB=BC=CD=DA Hence, ABCD is a rhombus.

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Description : ABCD is a rhombus and AB is produved to E and F such that AE=AB=BF prove that ED and FC are perpendicular to each other -Maths 9th

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P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Description : P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Last Answer : (i) In △ARB,P is the mid point of AB and PD || BR. ∴ D is a mid - point of AR [converse of mid - point theorem] ∴ AR = 2AD But BC = AD [opp sides of ||gm ABCD] Thus, AR = 2BC (ii) ∴ ABCD is a ... a mid - point of AR and DQ || AB ∴ Q is a mid point of BR [converse of mid - point theorem] ⇒ BR = 2BQ

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

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Last Answer : (i) In △ARB,P is the mid point of AB and PD || BR. ∴ D is a mid - point of AR [converse of mid - point theorem] ∴ AR = 2AD But BC = AD [opp sides of ||gm ABCD] Thus, AR = 2BC (ii) ∴ ABCD is a ... a mid - point of AR and DQ || AB ∴ Q is a mid point of BR [converse of mid - point theorem] ⇒ BR = 2BQ

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Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

Description : ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

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

ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

Description : ABCD is a parallelogram x and y are midpoints of BC and CD respectively.Prove that- Area of triangle axy =3/8 area of parallelogram ABCD -Maths 9th

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

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Description : ABCD is a parallelogram. P is a point on AD such that AP = 1/3 AD and Q is a point on BC such that CQ = 1/3 BC. Prove that AQCP is a parallelogram. -Maths 9th

Last Answer : answer:

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Description : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.

In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Description : In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP

In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Description : In the given figure, ABCD is a square. Side AB is produced to points P and Q in such a way that PA = AB = BQ. Prove that DQ = CP. -Maths 9th

Last Answer : In △PAD, ∠A = 90° and DA = PA = PB ⇒ ∠ADP = ∠APD = 90° / 2 = 45° Similarly, in △QBC, ∠B = 90° and BQ = BC = AB ⇒∠BCQ = ∠BQC = 90° / 2 = 45° In △PAD and △QBC , we have PA = QB [given] ∠A = ... [each = 90° + 45° = 135°] ⇒ △PDC = △QCD [by SAS congruence rule] ⇒ PC = QD or DQ = CP

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

Last Answer : Solution :-

Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus. -Maths 9th

Description : Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus. -Maths 9th

Last Answer : . Solution: (i) In ΔADC and ΔCBA, AD = CB (Opposite sides of a parallelogram) DC = BA (Opposite sides of a parallelogram) AC = CA (Common Side) , ΔADC ≅ ΔCBA [SSS congruency] Thus, ∠ACD = ∠CAB by ... are equal) Also, AB = BC = CD = DA (Opposite sides of a parallelogram) Thus, ABCD is a rhombus.

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Last Answer : Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC = BD and AC ⊥ BD. To prove PQRS is a square. Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,

P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Description : P, Q, R and S are respectively the mid-points of sides AB, BC, CD and DA of quadrilateral ABCD in which AC = BD and AC ⊥ BD. Prove that PQRS is a square. -Maths 9th

Last Answer : Given In quadrilateral ABCD, P, Q, R and S are the mid-points of the sides AB, BC, CD and DA, respectively. Also, AC = BD and AC ⊥ BD. To prove PQRS is a square. Proof Now, in ΔADC, S and R are the mid-points of the sides AD and DC respectively, then by mid-point theorem,

ABCD is a parallelogram in which BC is produced to E such that CE = BC . -Maths 9th

Description : ABCD is a parallelogram in which BC is produced to E such that CE = BC . -Maths 9th

Last Answer : In △ADF and △ECF , we have ∠ADF = ∠ECF [alt.int.∠s] AD = EC [ ∵ AD = BC and BC = EC] ∠DFA = ∠CFE [vert. opp. ∠s] ∴ By AAS congruence rule , △ADF ≅ △ECF ⇒ DF = CF [c.p.c.t.] ⇒ ar(△ADF) = ar(△ECF) ... 3 = 6 cm2 [∵ar(△DFB) = 3 cm2] Thus, ar(||gm ABCD) = 2 ar(△BDC) = 2 6 = 12 cm2

ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. -Maths 9th

Description : ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. -Maths 9th

Last Answer : According to question find the area of the parallelogram ABCD.

ABCD is a parallelogram in which BC is produced to E such that CE = BC . -Maths 9th

Description : ABCD is a parallelogram in which BC is produced to E such that CE = BC . -Maths 9th

Last Answer : In △ADF and △ECF , we have ∠ADF = ∠ECF [alt.int.∠s] AD = EC [ ∵ AD = BC and BC = EC] ∠DFA = ∠CFE [vert. opp. ∠s] ∴ By AAS congruence rule , △ADF ≅ △ECF ⇒ DF = CF [c.p.c.t.] ⇒ ar(△ADF) = ar(△ECF) ... 3 = 6 cm2 [∵ar(△DFB) = 3 cm2] Thus, ar(||gm ABCD) = 2 ar(△BDC) = 2 6 = 12 cm2

ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. -Maths 9th

Description : ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. -Maths 9th

Last Answer : According to question find the area of the parallelogram ABCD.

In Fig. 9.23, ABCD is a parallelogram in which BC is produced to E such A B that CE = BC. AE intersects CD at F. If area of △BDF = 3 cm2, find the area of parallelogram ABCD. -Maths 9th

Description : In Fig. 9.23, ABCD is a parallelogram in which BC is produced to E such A B that CE = BC. AE intersects CD at F. If area of △BDF = 3 cm2, find the area of parallelogram ABCD. -Maths 9th

Last Answer : Solution :-

ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. -Maths 9th

Description : ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. -Maths 9th

Last Answer : In ||gm ABCD , ar(△APC) = ar(△BCP) ---i) [∵ triangles on the same base and between the same parallels have equal area] Similarly, ar( △ADQ) = ar(△ADC) ---ii) Now, ar(△ADQ) - ar(△ADP) = ar(△ADC) - ar(△ADP) ... ) From (i) and (iii) , we have ar(△BCP) = ar(△DPQ) or ar( △BPC) = ar(△DPQ)

ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. -Maths 9th

Description : ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. -Maths 9th

Last Answer : In ||gm ABCD , ar(△APC) = ar(△BCP) ---i) [∵ triangles on the same base and between the same parallels have equal area] Similarly, ar( △ADQ) = ar(△ADC) ---ii) Now, ar(△ADQ) - ar(△ADP) = ar(△ADC) - ar(△ADP) ... ) From (i) and (iii) , we have ar(△BCP) = ar(△DPQ) or ar( △BPC) = ar(△DPQ)

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

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.

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

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.

ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus. -Maths 9th

Description : ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus. -Maths 9th

Last Answer : According to question altitude from D to side AB bisects AB. Find the angles of the rhombus.

ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus. -Maths 9th

Description : ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus. -Maths 9th

Last Answer : According to question altitude from D to side AB bisects AB. Find the angles of the rhombus.

ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus. -Maths 9th

Description : ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus. -Maths 9th

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ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Description : ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that (i) ∠A = ∠B (ii) ∠C = ∠D (iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.] -Maths 9th

Last Answer : ] Solution: To Construct: Draw a line through C parallel to DA intersecting AB produced at E. (i) CE = AD (Opposite sides of a parallelogram) AD = BC (Given) , BC = CE ⇒∠CBE = ∠CEB also, ∠A+∠CBE = ... BC (Given) , ΔABC ≅ ΔBAD [SAS congruency] (iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.

In Fig. 7.19, AD and BC are equal perpendicular to a line segment AB. Show that CD bisects AB. -Maths 9th

Description : In Fig. 7.19, AD and BC are equal perpendicular to a line segment AB. Show that CD bisects AB. -Maths 9th

Last Answer : Solution :-

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

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In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Description : In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Last Answer : Given In figure AB || DE and AC || DF, also AB = DE and AC = DF To prove BC ||EF and BC = EF Proof In quadrilateral ABED, AB||DE and AB = DE So, ABED is a parallelogram. AD || BE and AD = BE Now, ... = CF and BE||CF [from Eq. (iii)] So, BCFE is a parallelogram. BC = EF and BC|| EF . Hence proved.

In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Description : In figure, AB || DE, AB = DE, AC|| DF and AC = OF. Prove that BC || EF and BC = EF. -Maths 9th

Last Answer : Given In figure AB || DE and AC || DF, also AB = DE and AC = DF To prove BC ||EF and BC = EF Proof In quadrilateral ABED, AB||DE and AB = DE So, ABED is a parallelogram. AD || BE and AD = BE Now, ... = CF and BE||CF [from Eq. (iii)] So, BCFE is a parallelogram. BC = EF and BC|| EF . Hence proved.

In the given figure, O is the centre of the circle. The radius OP bisects a rectangle ABCD at right angles. -Maths 9th

Description : In the given figure, O is the centre of the circle. The radius OP bisects a rectangle ABCD at right angles. -Maths 9th

Last Answer : answer:

In the adjoining figure, points A, B, C and D lie on a circle. AD = 24 and BC = 12. -Maths 9th

Description : In the adjoining figure, points A, B, C and D lie on a circle. AD = 24 and BC = 12. -Maths 9th

Last Answer : AD = 24, BC = 12. In ΔCBE and ΔADE, ∠CBA = ∠CDA, ∠BCE = ∠DAE (Angles in the same segment are equal) ∠BEC = ∠DEA (vertical opposite angles are equal) ⇒ ΔBCE and ΔDEA are similar Δs with sides in the ratio 1 : 2. ∴ Ratio of areas = Ratio of square of sides = 12 : 22 = 1 : 4

ABCD is a cyclic quadrilateral such that AC ⊥ BD. AC meet BD at E. Prove that EA^2 + EB^2 + EC^2 + ED^2 = 4R^2, -Maths 9th

Description : ABCD is a cyclic quadrilateral such that AC ⊥ BD. AC meet BD at E. Prove that EA^2 + EB^2 + EC^2 + ED^2 = 4R^2, -Maths 9th

Last Answer : answer:

ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. -Maths 9th

Description : ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. -Maths 9th

Last Answer : Given ABCD is a parallelogram. A circle whose centre O passes through A, B is so drawn that it intersect AD at P and BC at Q To prove Points P, Q, C and D are con-cyclic. Construction Join PQ ... Thus, the quadrilateral QCDP is cyclic. So, the points P, Q, C and D are con-cyclic. Hence proved.

ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. -Maths 9th

Description : ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. -Maths 9th

Last Answer : Given ABCD is a parallelogram. A circle whose centre O passes through A, B is so drawn that it intersect AD at P and BC at Q To prove Points P, Q, C and D are con-cyclic. Construction Join PQ ... Thus, the quadrilateral QCDP is cyclic. So, the points P, Q, C and D are con-cyclic. Hence proved.

A point E is taken on the side BC of a parallelogram ABCD. -Maths 9th

Description : A point E is taken on the side BC of a parallelogram ABCD. -Maths 9th

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A point E is taken on the side BC of a parallelogram ABCD. -Maths 9th

Description : A point E is taken on the side BC of a parallelogram ABCD. -Maths 9th

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The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm

The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th

Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm

P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Description : P and O are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. -Maths 9th

Last Answer : According to question PQ passes through the point of intersection O of its diagonals AC and BD.