Advertisements
Advertisements
Question
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
Solution
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°
RELATED QUESTIONS
Let Abc Be an Isosceles Triangle in Which Ab = Ac. If D, E, F Be the Mid-points of the Sides Bc, Ca and a B Respectively, Show that the Segment Ad and Ef Bisect Each Other at Right Angles.
ABC is a triang D is a point on AB such that AD = `1/4` AB and E is a point on AC such that AE = `1/4` AC. Prove that DE = `1/4` BC.
Fill in the blank to make the following statement correct
The triangle formed by joining the mid-points of the sides of an isosceles triangle is
Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus.
In ∆ABC, E is the mid-point of the median AD, and BE produced meets side AC at point Q.
Show that BE: EQ = 3: 1.
ABCD is a quadrilateral in which AD = BC. E, F, G and H are the mid-points of AB, BD, CD and Ac respectively. Prove that EFGH is a rhombus.
Use the following figure to find:
(i) BC, if AB = 7.2 cm.
(ii) GE, if FE = 4 cm.
(iii) AE, if BD = 4.1 cm
(iv) DF, if CG = 11 cm.
In trapezium ABCD, sides AB and DC are parallel to each other. E is mid-point of AD and F is mid-point of BC.
Prove that: AB + DC = 2EF.
In triangle ABC; M is mid-point of AB, N is mid-point of AC and D is any point in base BC. Use the intercept Theorem to show that MN bisects AD.
In ΔABC, AB = 12 cm and AC = 9 cm. If M is the mid-point of AB and a straight line through M parallel to AC cuts BC in N, what is the length of MN?
In ΔABC, D, E, F are the midpoints of BC, CA and AB respectively. Find FE, if BC = 14 cm
Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a square is also a square.
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
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: ∠EFG = 90°
The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if ______.
The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is ______.
The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if, ______.
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.
