English
Maharashtra State BoardSSC (English Medium) 9th Standard

In the given figure, ΔABC is an equilateral traingle. Points F, D and E are midpoints of side AB, side BC, side AC respectively. Show that ΔFED is an equilateral traingle. - Geometry

Advertisements
Advertisements

Question

In the given figure, ΔABC is an equilateral traingle. Points F, D and E are midpoints of side AB, side BC, side AC respectively. Show that ΔFED is an equilateral traingle.

Sum
Advertisements

Solution

Given: ∆ABC is an equilateral triangle and D, E and F are mid-points of BC, AC and AB respectively.

To prove: ∆FED is an equilateral triangle.

Proof:

In ΔABC,

Points F and E are the midpoints of sides AB and AC respectively.      ...(Given)

∴ FE = `1/2` BC       ...(From midpoint theorem) ...(i)

In ΔABC,

Points D and E are the midpoints of sides BC and AC respectively.     ...(Given)

∴ DE = `1/2` AB      ...(From midpoint theorem)   ...(ii)

In ΔABC,

Points D and F are the midpoints of sides BC and AB respectively.     ...(Given)

∴ DF = `1/2` AC       ...(From midpoint theorem) ...(iii)

Now, ΔABC is an equilateral triangle.

∴ BC = AB = AC      ...(Sides of equilateral triangle)

∴ `1/2` BC = `1/2` AB = `1/2` AC     ...(Multiplying both sides by `1 /2`)

∴ FE = DE = DF       ...[From (i), (ii) and (iii)]

∴ ΔFED is an equilateral triangle.

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Quadrilaterals - Practice Set 5.5 [Page 73]

APPEARS IN

Balbharati Mathematics 2 [English] Standard 9 Maharashtra State Board
Chapter 5 Quadrilaterals
Practice Set 5.5 | Q 3 | Page 73

RELATED QUESTIONS

In below fig. ABCD is a parallelogram and E is the mid-point of side B If DE and AB when produced meet at F, prove that AF = 2AB.


In a ΔABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC
intersects FE at Q. Prove that AQ = QP.


In a ΔABC, BM and CN are perpendiculars from B and C respectively on any line passing
through A. If L is the mid-point of BC, prove that ML = NL.


ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively
intersecting at P, Q and R. Prove that the perimeter of ΔPQR is double the perimeter of
ΔABC


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.


Show that the line segments joining the mid-points of the opposite sides of a quadrilateral
bisect each other.


In the given figure, seg PD is a median of ΔPQR. Point T is the mid point of seg PD. Produced QT intersects PR at M. Show that `"PM"/"PR" = 1/3`.

[Hint: DN || QM]


In a triangle ABC, AD is a median and E is mid-point of median AD. A line through B and E meets AC at point F.

Prove that: AC = 3AF.


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 the given figure, AD and CE are medians and DF // CE.
Prove that: FB = `1/4` AB.


In ΔABC, D, E, F are the midpoints of BC, CA and AB respectively. Find FE, if BC = 14 cm


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


AD is a median of side BC of ABC. E is the midpoint of AD. BE is joined and produced to meet AC at F. Prove that AF: AC = 1 : 3.


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.


In the given figure, T is the midpoint of QR. Side PR of ΔPQR is extended to S such that R divides PS in the ratio 2:1. TV and WR are drawn parallel to PQ. Prove that T divides SU in the ratio 2:1 and WR = `(1)/(4)"PQ"`.


E is the mid-point of the side AD of the trapezium ABCD with AB || DC. A line through E drawn parallel to AB intersect BC at F. Show that F is the mid-point of BC. [Hint: Join AC]


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×