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
ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see the given figure). AC is a diagonal. Show that:
- SR || AC and SR = `1/2AC`
- PQ = SR
- PQRS is a parallelogram.

ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
- D is the mid-point of AC
- MD ⊥ AC
- CM = MA = `1/2AB`
In Fig. below, M, N and P are the mid-points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5 cm and MP = 2.5 cm, calculate BC, AB and AC.

The diagonals of a quadrilateral intersect at right angles. Prove that the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral is rectangle.
In triangle ABC, angle B is obtuse. D and E are mid-points of sides AB and BC respectively and F is a point on side AC such that EF is parallel to AB. Show that BEFD is a parallelogram.
In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively.
Prove that:
(i) Triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.
In ΔABC, D, E, F are the midpoints of BC, CA and AB respectively. Find DE, if AB = 8 cm
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.
In ΔABC, BE and CF are medians. P is a point on BE produced such that BE = EP and Q is a point on CF produced such that CF = FQ. Prove that: QAP is a straight line.
Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.
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: 3DF = EF
In a parallelogram ABCD, E and F are the midpoints of the sides AB and CD respectively. The line segments AF and BF meet the line segments DE and CE at points G and H respectively Prove that: EGFH is a parallelogram.
In ΔABC, D and E are the midpoints of the sides AB and AC respectively. F is any point on the side BC. If DE intersects AF at P show that DP = PE.
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
D and E are the mid-points of the sides AB and AC of ∆ABC and O is any point on side BC. O is joined to A. If P and Q are the mid-points of OB and OC respectively, then DEQP is ______.
The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if, ______.
D, E and F are the mid-points of the sides BC, CA and AB, respectively of an equilateral triangle ABC. Show that ∆DEF is also an equilateral triangle.
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. Prove that PQRS is a rhombus.
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).
[Hint: Join BE and produce it to meet CD produced at G.]
