Advertisements
Advertisements
प्रश्न
In triangle ABC, D and E are points on side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meets side BC at points M and N respectively. Prove that: BM = MN = NC.
Advertisements
उत्तर
The figure is shown below
AD = DE = EB
In ΔAEG
AD = DE & DF || EG
Using mid point theorem
F is midpoint of AG
∴ AF = FG ...(1)
DF || EG || BC and DE = BE,
∴ FG = GC ...(2)
(1), (2) we get
AF = GF = GC
Similarly since GN || FM || AB
∴ BM = MN = NC
Hence, proved.
APPEARS IN
संबंधित प्रश्न
Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
In Fig. below, triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm and D,
E are the mid-points of the sides AB and AC respectively, calculate
(i) The length of BC (ii) The area of ΔADE.
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.
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
In the adjacent figure, `square`ABCD is a trapezium AB || DC. Points M and N are midpoints of diagonal AC and DB respectively then prove that MN || AB.
The side AC of a triangle ABC is produced to point E so that CE = 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 meet AC at point P and EF at point R respectively.
Prove that:
- 3DF = EF
- 4CR = AB
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, 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.
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.
