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Syeda
answered
1 year(s)
ago

AD is a median of triangle ABC and E is the mid-point of AD

Answer.

Given AD is the median of ΔABC and E is the midpoint of AD

Through D, draw DG || BF

In ΔADG, E is the midpoint of AD and EF || DG

By converse of midpoint theorem w

- 3 Answers
- 6 Likes
- 68756 Views

Agam
answered
1 year(s)
, 7 month(s) ago

prove that bisectors of angles of parallelogram form a rectangle

Given: ABCD is a parallelogram. AE bisects ∠BAD. BF bisects ∠ABC. CG bisects ∠BCD and DH bisects ∠ADC

To prove: LKJI is a rectangle

∠BAD + ∠ABC = 180° because adjacent angles of a paralle...

- 3 Answers
- 7 Likes
- 61619 Views

Syeda
answered
1 year(s)
ago

Prove that the sum of the four angles of a quadrilateral is 360 degree?

Consider a quadrilateral PQRS.

Join QS.

**To prove:** ∠P + ∠Q + ∠R + ∠S = 360º

**Proof:**

Consider triangle PQS, we have,

⇒ ∠P + ∠PQS

...- 2 Answers
- 2 Likes
- 41486 Views

Syeda
answered
1 year(s)
ago

ABCD is a parallelogram.X and y are mid points of BC and CD

__ Construction :-__ Join points B and D.

Since X and Y are the mid points of sides BC and CD respectively,

In

- 3 Answers
- 0 Like
- 39780 Views

Himanshu
answered
2 year(s)
, 6 month(s) ago

Show that the quadrilateral formed by joining the mid-points of its sides is a rectangle.

Given AC,BD are diagonals of a quadrilateral ABCD are perpendicular.

P,Q,R and S are the mid points of AB,BC, CD and AD respectively.

Proof:

In ΔABC, P an...

- 2 Answers
- 3 Likes
- 38513 Views

Agam
answered
1 year(s)
, 7 month(s) ago

Show that the line segment joining the midpoint of the opposite side of a quadrilateral bisect each other.

In ΔADC, S and R are the midpoints of AD and DC respectively.

Recall that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and half of i...

- 2 Answers
- 6 Likes
- 36408 Views

Syeda
answered
1 year(s)
ago

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

ABCD is a trapezium in which AB || DC and E, F are mid points of AD, BC respectively.

Join CE and produce it to meet BA produced at G.

In ΔEDC and ΔEAG,

<...

- 2 Answers
- 0 Like
- 35362 Views

Raghunath
answered
4 year(s)
, 5 month(s) ago

Show that the diagonals of a square are equal and bisect each other at right angles

Sol:

Given that ABCD is a square.

To prove : AC = BD and AC and BD bisect each other at right angles.

Proof:

(i) In a Δ ABC and Δ BAD,

...

- 1 Answer
- 5 Likes
- 34013 Views

Kishore
answered
7 year(s)
, 6 month(s) ago

Line segment joining the midpoints of the diagonals of a trapezium

Let E and F are midpoints of the diagonals AC and BD of trapezium ABCD respectively.

Draw DE and produce it to meet AB at G.

Consider DAEG and DCED

∠AEG = ∠CED (vertically

...Asked by Rajiv singh

Dec 29

- 1 Answer
- 2 Likes
- 28481 Views

Raghunath
answered
4 year(s)
, 4 month(s) ago

In a quadrilateral ABCD, show that OA+OB+OC+OD> AC+BD

**Sol:**

**Given :** ABCD is a quadrilateral. O is a point inside the quadrilateral ABCD.

**To prove : **OA + OB + OC + OD > AC + BD

- 1 Answer
- 6 Likes
- 26154 Views

Akhilesh
answered
4 year(s)
, 2 month(s) ago

Find the cost of white washing the walls of the room.

total area for white washing = lateral surface area + l*b

= 2*(l+b)*h + l*b

= 2*(5+4)*3 + 5*4...

- 2 Answers
- 0 Like
- 20589 Views

Raghunath
answered
4 year(s)
, 6 month(s) ago

BM and CN are perpendiculars to a line passing throughthe vertex A of a triangle ABC.If L is the mid-point of BC, prove that LM=LN.

Sol:

Given: In a ΔABC*l* is a straight line passing through the vertex A . BM ⊥ *l* and CN ⊥ *l*. L is the mid point of BC.

To prove**:** LM = LN

C

...Asked by Shatoddruh

Dec 6

- 1 Answer
- 2 Likes
- 17656 Views

Agam
answered
1 year(s)
, 10 month(s) ago

2 parallel lines l and m intersect by transversal p.

Given l||m and p is the transversal

To prove: PQRS is a rectangle

Proof:

RS, PS, PQ and RQ are bisectors of interior angles formed by the transversal with the parallel lines....

- 2 Answers
- 3 Likes
- 16645 Views

Kishore
answered
5 year(s)
, 7 month(s) ago

In a parallelogram ABCD, E and F are the midpoints

Given ABCD is a parallelogram

Hence AB || CD

⇒ AE || FC

Also AB = CD (Opposite sides of parallelogram ABCD)

⇒ AE = FC (Since E and F are midpoints of AB and CD)

In...

- 1 Answer
- 0 Like
- 14327 Views

Raghunath
answered
4 year(s)
, 3 month(s) ago

ABCD is a cyclic quadrilateral whose diagonal intersects at E. if ∠BAC = 30° and ∠DBC = 70° . find∠BCD. if AB=BC find ∠ECD.

ABCD is a cyclic quadrilateral whose diagonal intersect at E.

∠CBD = ∠CAD (Angles in the same segment are equal)

∠CAD = 70°

∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°

∠BCD +

...- 1 Answer
- 0 Like
- 13831 Views

Ramesh
answered
4 year(s)
, 8 month(s) ago

Prove that (i) AE = BE (II) ∠DAE = 15°.

Given : ABCD is a square and EDC is an equilateral triangle. AD = BC, DE = CE

To Prove : i) AE = BE

ii) ∠DAE = 15°

Construction : Join A to E and B to E. <...

- 2 Answers
- 2 Likes
- 13518 Views

Agam
answered
1 year(s)
, 10 month(s) ago

ABCD is a trapezium in which AB is parallel to CD.

Given in trapezium ABCD, AB||CD and AD = BC

Draw perpendiculars DP and CQ on AB

Consider triangles APD and BQC

∠P = ∠Q = 90°

DP = CQ [Distance between parallel sides is sa...

- 2 Answers
- 1 Like
- 13263 Views

Subhashri
commented
3 year(s)
, 3 month(s) ago

In the below figure, ABCD is a parallelogram and DAB = 60 . If the bisector AP and BP of angles A and B respectively meet P on CD. Prove that P is the midpoint of CD.

Since AB∥DC and Ap is a transversal,

∠PAB=∠DPA (Alternate angles)

but,∠PAB=∠DAP (Given)

∠DPA=∠DAP=30º

so, DP=DA (Isosceles triangle Property)

similarl...

- 2 Answers
- 1 Like
- 12502 Views

Ramesh
answered
4 year(s)
, 5 month(s) ago

Prove that AF=2AB.

Sol:

ABCD is a parallelogram. E is the midpoint of BC. So, BE = CE.

DE produced meets the AB produced at F.

Consider the triangles CDE and BFE.

BE = CE [Given]...

- 1 Answer
- 1 Like
- 9641 Views

Kishore
answered
5 year(s)
, 3 month(s) ago

Find the angles of the parallelogram.

Given ABCD is a parallelogram in which DP⊥AB and AQ ⊥BC.

Given ∠PDQ = 60°

In quad. DPBQ, by angle sum property we have

∠PDQ + ∠DPB + ∠B + ∠BQD = 360°

60° + 90° + ∠B + 90°...

Asked by Kamal kishore

Mar 17

- 1 Answer
- 3 Likes
- 8536 Views

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