मराठी

ABCD is a rhombus. EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced, meet at right angles.

Advertisements
Advertisements

प्रश्न

ABCD is a rhombus. EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced, meet at right angles.

सिद्धांत
Advertisements

उत्तर


We know that the diagonals of a rhombus are perpendicular bisectors of each other

∴ OA = OC

OB = OD

∠AOD = ∠COD = 90°

And ∠AOB = ∠COB = 90°

In ΔBDE, A and O are midpoints of BE and BD, respectively

OA || DE

OC || DG

In ΔCFA, B and O are midpoints of AF and AC, respectively

∴ OB || CF

OD || GC

Thus, in quadrilateral DOCG, we have

OC || DG and OD || GC

⇒ DOCG is a parallelogram

∠DGC = ∠DOC

∠DGC = 90°

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Quadrilaterals - Exercise 13.3 [पृष्ठ ४३]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 9
पाठ 13 Quadrilaterals
Exercise 13.3 | Q 8 | पृष्ठ ४३
नूतन Mathematics [English] Class 9 ICSE
पाठ 9 Mid-point Theorem
Exercise 9A | Q 16. | पृष्ठ १९६

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see the given figure). Show that the line segments AF and EC trisect the diagonal BD.


In a ∆ABC, D, E and F are, respectively, the mid-points of BC, CA and AB. If the lengths of side AB, BC and CA are 7 cm, 8 cm and 9 cm, respectively, find the perimeter of ∆DEF.


In a ΔABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC
intersects FE at Q. Prove that AQ = QP.


In below Fig, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = `1/4` AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.


Fill in the blank to make the following statement correct:

The triangle formed by joining the mid-points of the sides of a right triangle is            


In the Figure, `square`ABCD is a trapezium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively. Then prove that, PQ || AB and PQ = `1/2 ("AB" + "DC")`.


In a triangle ABC, AD is a median and E is mid-point of median AD. A line through B and E meets AC at point F.

Prove that: AC = 3AF.


In trapezium ABCD, AB is parallel to DC; P and Q are the mid-points of AD and BC respectively. BP produced meets CD produced at point E.

Prove that:

  1. Point P bisects BE,
  2. PQ is parallel to AB.

In triangle ABC ; D and E are mid-points of the sides AB and AC respectively. Through E, a straight line is drawn parallel to AB to meet BC at F.
Prove that BDEF is a parallelogram. If AB = 16 cm, AC = 12 cm and BC = 18 cm,
find the perimeter of the parallelogram BDEF.


In parallelogram PQRS, L is mid-point of side SR and SN is drawn parallel to LQ which meets RQ produced at N and cuts side PQ at M. Prove that M is the mid-point of PQ.


D, E and F are the mid-points of the sides AB, BC and CA of an isosceles ΔABC in which AB = BC. Prove that ΔDEF is also isosceles.


In the given figure, ABCD is a trapezium. P and Q are the midpoints of non-parallel side AD and BC respectively. Find: DC, if AB = 20 cm and PQ = 14 cm


Side AC of a ABC is produced to point E so that CE = `(1)/(2)"AC"`. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meets AC at point P and EF at point R respectively. Prove that: 4CR = AB.


ABCD is a kite in which BC = CD, AB = AD. E, F and G are the mid-points of CD, BC and AB respectively. Prove that: The line drawn through G and parallel to FE and bisects DA.


In ΔABC, X is the mid-point of AB, and Y is the mid-point of AC. BY and CX are produced and meet the straight line through A parallel to BC at P and Q respectively. Prove AP = AQ.


In ΔABC, the medians BE and CD are produced to the points P and Q respectively such that BE = EP and CD = DQ. Prove that: A is the mid-point of PQ.


In ΔABC, D and E are the midpoints of the sides AB and BC respectively. F is any point on the side AC. Also, EF is parallel to AB. Prove that BFED is a parallelogram.

Remark: Figure is incorrect in Question


In the given figure, PS = 3RS. M is the midpoint of QR. If TR || MN || QP, then prove that:

ST = `(1)/(3)"LS"`


E is the mid-point of the side AD of the trapezium ABCD with AB || DC. A line through E drawn parallel to AB intersect BC at F. Show that F is the mid-point of BC. [Hint: Join AC]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×