ABCD is a parallelogram. The diagonals AC and BD intersect at the point O. If E, F, G and H are the mid-points of AO, DO, CO and BO respectively -Maths 9th

ABCD is a parallelogram. The diagonals AC and BD intersect at the point O. If E, F, G and H are the mid-points of AO, DO, CO and BO respectively -Maths 9th
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The diagonals AC and BD of parallelogram ABCD intersect at the point O. -Maths 9th

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

Last Answer : ABCD is a parallelogram . ∴ AD | | BC ⇒ ∠ACB = ∠DAC = 34° Now, ∠AOB is an exterior angle of △BOC ∴ ∠OBC + OCB = ∠AOB [∵ ext ∠ = sum of two int. opp. ∠S] ⇒ ∠OBC + 34° = 75° ⇒ ∠OBC = 75° - 34° = 41° or ∠DBC = 41°

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

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

Last Answer : ABCD is a parallelogram . ∴ AD | | BC ⇒ ∠ACB = ∠DAC = 34° Now, ∠AOB is an exterior angle of △BOC ∴ ∠OBC + OCB = ∠AOB [∵ ext ∠ = sum of two int. opp. ∠S] ⇒ ∠OBC + 34° = 75° ⇒ ∠OBC = 75° - 34° = 41° or ∠DBC = 41°

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

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

Last Answer : According to parallelogram ABCD intersect each other at O. If OA = 3 cm and OD = 2 cm, determine the lengths of AC and BD.

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

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

Last Answer : According to parallelogram ABCD intersect each other at O. If OA = 3 cm and OD = 2 cm, determine the lengths of AC and BD.

The diagonals AC and BD of a parallelogram ABCD intersect each other at the point 0. -Maths 9th

Description : The diagonals AC and BD of a parallelogram ABCD intersect each other at the point 0. -Maths 9th

Last Answer : According to question parallelogram ABCD intersect each other at the point 0. If ∠DAC = 32° and ∠AOB = 70°.

The diagonals AC and BD of a parallelogram ABCD intersect each other at the point 0. -Maths 9th

Description : The diagonals AC and BD of a parallelogram ABCD intersect each other at the point 0. -Maths 9th

Last Answer : According to question parallelogram ABCD intersect each other at the point 0. If ∠DAC = 32° and ∠AOB = 70°.

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)

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)

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.

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.

The diagonals AC and BD of a cyclic quadrilateral ABCD intersect at P. Let O be the circumcentre of ∆APB and H be the orthocentre -Maths 9th

Description : The diagonals AC and BD of a cyclic quadrilateral ABCD intersect at P. Let O be the circumcentre of ∆APB and H be the orthocentre -Maths 9th

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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.

Diagonals AC and BC of parallelogram ABCD Intersect at point O. Angle BOC=90° and BDC=50°.find angle OAB. -Maths 9th

Description : Diagonals AC and BC of parallelogram ABCD Intersect at point O. Angle BOC=90° and BDC=50°.find angle OAB. -Maths 9th

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Diagonals AC and BC of parallelogram ABCD Intersect at point O. Angle BOC=90° and BDC=50°.find angle OAB. -Maths 9th

Description : Diagonals AC and BC of parallelogram ABCD Intersect at point O. Angle BOC=90° and BDC=50°.find angle OAB. -Maths 9th

Last Answer : Its given ABCD is a IIgram and AC and BD are its diagonals intersecting at point O. . Given : angle BOC = 900 angle BDC = 500 To find : angle OAB Answer : i) ...

In trapezium ABCD, AB|| DC and diagonals AC and BD intersect at O. If area of triangle AOD is 30cm square , find the area of triangle BOC -Maths 9th

Description : In trapezium ABCD, AB|| DC and diagonals AC and BD intersect at O. If area of triangle AOD is 30cm square , find the area of triangle BOC -Maths 9th

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.

In trapezium ABCD, AB|| DC and diagonals AC and BD intersect at O. If area of triangle AOD is 30cm square , find the area of triangle BOC -Maths 9th

Description : In trapezium ABCD, AB|| DC and diagonals AC and BD intersect at O. If area of triangle AOD is 30cm square , find the area of triangle BOC -Maths 9th

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.

A trapezium ABCD in which AB || CD is inscribed in a circle with centre O. Suppose the diagonals AC and BD of the trapezium intersect at M -Maths 9th

Description : A trapezium ABCD in which AB || CD is inscribed in a circle with centre O. Suppose the diagonals AC and BD of the trapezium intersect at M -Maths 9th

Last Answer : answer:

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)

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)

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.

The diagonals of a parallelogram ABCD intersect at a point O. -Maths 9th

Description : The diagonals of a parallelogram ABCD intersect at a point O. -Maths 9th

Last Answer : According to question PQ divides the parallelogram into two parts of equal area.

The diagonals of a parallelogram ABCD intersect at a point O. -Maths 9th

Description : The diagonals of a parallelogram ABCD intersect at a point O. -Maths 9th

Last Answer : According to question PQ divides the parallelogram into two parts of equal area.

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

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

Last Answer : Given In a quadrilateral ABCD, P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. Also, AC = BD To prove PQRS 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 the sides AB, BC, CD and DA of a quadrilateral ABCD in which AC = BD. -Maths 9th

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

Last Answer : Given In a quadrilateral ABCD, P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. Also, AC = BD To prove PQRS 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,

Let ABCD be a quadrilateral. Let X and Y be the mid-points of AC and BD respectively and the lines through X and Y respectively -Maths 9th

Description : Let ABCD be a quadrilateral. Let X and Y be the mid-points of AC and BD respectively and the lines through X and Y respectively -Maths 9th

Last Answer : answer:

The middle points of the parallel sides AB and CD of a parallelogram ABCD are P and Q respectively. If AQ and CP divide the diagonal BD -Maths 9th

Description : The middle points of the parallel sides AB and CD of a parallelogram ABCD are P and Q respectively. If AQ and CP divide the diagonal BD -Maths 9th

Last Answer : answer:

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.

BD is one of the diagonals of a quadrilateral ABCD. AM and CN are the perpendiculars from A and C respectively on BD . -Maths 9th

Description : BD is one of the diagonals of a quadrilateral ABCD. AM and CN are the perpendiculars from A and C respectively on BD . -Maths 9th

Last Answer : We know that area of a triangle = 1/2 × base × altitude ∴ ar(△ABD) = 1/2 × BD × AM and ar(△BCD) = 1/2 BD × CN Now, ar(quad. ABCD) = ar(△ABD) + ar(△BCD) = 1/2 × BD × AM + 1/2 × BD × CN = 1/2 × BD × (AM + CN)

BD is one of the diagonals of a quadrilateral ABCD. AM and CN are the perpendiculars from A and C respectively on BD . -Maths 9th

Description : BD is one of the diagonals of a quadrilateral ABCD. AM and CN are the perpendiculars from A and C respectively on BD . -Maths 9th

Last Answer : We know that area of a triangle = 1/2 × base × altitude ∴ ar(△ABD) = 1/2 × BD × AM and ar(△BCD) = 1/2 BD × CN Now, ar(quad. ABCD) = ar(△ABD) + ar(△BCD) = 1/2 × BD × AM + 1/2 × BD × CN = 1/2 × BD × (AM + CN)

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.

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2

If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Description : If P,Q,R,S are respectively the mid - points of the sides of a parallelogram ABCD, if ar(||gm PQRS) = 32.5cm2 , then find ar(||gm ABCD). -Maths 9th

Last Answer : Join PR. ∵ △PSR and ||gm APRD are on the same base and between same parallel lines. ar(△PSR) = 1/2 ar(||gm APRD) Similarly, ar(△PQR) = 1/2 ar(||gm PBCR) ar(△PQRS) = ar(△PSR) + △(PQR) = 1/2 ar(||gm APRD) + 1 ... |gm PBCR) = 1/2 ar(||gm ABCD) ⇒ ar(||gm ABCD) = 2 ar(||gm PQRS) = 2 32.5 = 65cm2

ABCD is a parallelogram and O is the point of intersection of its diagonals. -Maths 9th

Description : ABCD is a parallelogram and O is the point of intersection of its diagonals. -Maths 9th

Last Answer : Here, ABCD is a parallelogram in which its diagonals AC and BD intersect each other in O. ∴ O is the mid - point of AC as well as BD. Now, in △ADB , AO is its median ∴ ar(△ADB) = 2 ar(△AOD) [ ∵ median ... AB and lie between same parallel AB and CD . ∴ ar(ABCD) = 2 ar(△ADB) = 2 8 = 16 cm2

ABCD is a parallelogram and O is the point of intersection of its diagonals. -Maths 9th

Description : ABCD is a parallelogram and O is the point of intersection of its diagonals. -Maths 9th

Last Answer : Here, ABCD is a parallelogram in which its diagonals AC and BD intersect each other in O. ∴ O is the mid - point of AC as well as BD. Now, in △ADB , AO is its median ∴ ar(△ADB) = 2 ar(△AOD) [ ∵ median ... AB and lie between same parallel AB and CD . ∴ ar(ABCD) = 2 ar(△ADB) = 2 8 = 16 cm2

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)

ABCD is a square. E and F are respectively the mid - points of BC and CD. If R is the mid point of EF. -Maths 9th

Description : ABCD is a square. E and F are respectively the mid - points of BC and CD. If R is the mid point of EF. -Maths 9th

Last Answer : Since R is the mid point of EF . ∴ AR is the median in △AEF. As, a median of a triangle divides it into two triangles of equal area . ∴ ar(△AER) = ar(△AFR)

ABCD is a square. E and F are respectively the mid - points of BC and CD. If R is the mid point of EF. -Maths 9th

Description : ABCD is a square. E and F are respectively the mid - points of BC and CD. If R is the mid point of EF. -Maths 9th

Last Answer : Since R is the mid point of EF . ∴ AR is the median in △AEF. As, a median of a triangle divides it into two triangles of equal area . ∴ ar(△AER) = ar(△AFR)

In a trapezium ABCD, AB is parallel to CD and the diagonals intersect each other at O. In this case, the ratio OA/OC is equal to: -Maths 9th

Description : In a trapezium ABCD, AB is parallel to CD and the diagonals intersect each other at O. In this case, the ratio OA/OC is equal to: -Maths 9th

Last Answer : answer:

ABCD is a trapezium with AB and CD as parallel sides. The diagonals intersect at O. The area of the triangle ABO is p and that of triangle CDO is q. -Maths 9th

Description : ABCD is a trapezium with AB and CD as parallel sides. The diagonals intersect at O. The area of the triangle ABO is p and that of triangle CDO is q. -Maths 9th

Last Answer : answer:

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.

Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Last Answer : According to question parallelogram ABCD such that AP = CQ.

Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Last Answer : According to question parallelogram ABCD such that AP = CQ.

In Fig. 8.40, points M and N are taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AM = CN. Show that AC and MN bisect each other. -Maths 9th

Description : In Fig. 8.40, points M and N are taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AM = CN. Show that AC and MN bisect each other. -Maths 9th

Last Answer : Solution :-

In Fig. 8.37, ABCD is a parallelogram and P, Q are the points on the diagonal BD such that BQ = DP. Show what APCQ is a parallelogram. -Maths 9th

Description : In Fig. 8.37, ABCD is a parallelogram and P, Q are the points on the diagonal BD such that BQ = DP. Show what APCQ is a parallelogram. -Maths 9th

Last Answer : Solution :-

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

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.