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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

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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

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.

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Chapter 5: Quadrilaterals - Practice Set 5.5 [Page 73]

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

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