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.
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°
APPEARS IN
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:
- Point P bisects BE,
- 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]
