Advertisements
Advertisements
प्रश्न
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
Advertisements
उत्तर
From the figure EF || AB and E is the midpoint of BC.
Therefore, F is the midpoint of AC.
Here EF || BD, EF = BD as D is the midpoint of AB.
BE || DF, BE = DF as E is the midpoint of BC.
Therefore BEFD is a parallelogram.
Remark: Figure modified.
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.
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.
In Fig. below, BE ⊥ AC. AD is any line from A to BC intersecting BE in H. P, Q and R are
respectively the mid-points of AH, AB and BC. Prove that ∠PQR = 90°.
BM and CN are perpendiculars to a line passing through the vertex A of a triangle ABC. If
L is the mid-point of BC, prove that LM = LN.
L and M are the mid-point of sides AB and DC respectively of parallelogram ABCD. Prove that segments DL and BM trisect diagonal AC.
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.
ΔABC is an isosceles triangle with AB = AC. D, E and F are the mid-points of BC, AB and AC respectively. Prove that the line segment AD is perpendicular to EF and is bisected by it.
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, D, E and F are the midpoints of AB, BC and AC.
Show that AE and DF bisect each other.
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: ΔHEB ≅ ΔHFC
