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
ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.
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.
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
The figure, given below, shows a trapezium ABCD. M and N are the mid-point of the non-parallel sides AD and BC respectively. Find:
- MN, if AB = 11 cm and DC = 8 cm.
- AB, if DC = 20 cm and MN = 27 cm.
- DC, if MN = 15 cm and AB = 23 cm.
In triangle ABC, AD is the median and DE, drawn parallel to side BA, meets AC at point E.
Show that BE is also a median.
A parallelogram ABCD has P the mid-point of Dc and Q a point of Ac such that
CQ = `[1]/[4]`AC. PQ produced meets BC at R.
Prove that
(i)R is the midpoint of BC
(ii) PR = `[1]/[2]` DB
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.
ABCD is a parallelogram.E is the mid-point of CD and P is a point on AC such that PC = `(1)/(4)"AC"`. EP produced meets BC at F. Prove that: F is the mid-point of BC.
ABCD is a parallelogram.E is the mid-point of CD and P is a point on AC such that PC = `(1)/(4)"AC"`. EP produced meets BC at F. Prove that: 2EF = BD.
