Question

D e and f are respectively the midpoints of side bc ca and a b of triangle abc show that area of triangle d e f is equal to 1 by 4 of triangle abc

Answer 1

Step-by-step explanation:

Given that - In ∆ABC

1. D is the midpoint of BC
2. F is the midpoint of AB
3. E is the midpoint of AC

To prove - ∆DEF = 1/4 ∆ABC

ans -

consider ΔABC

E and F are midpoints of the sides AB and AC

The line joining the midpoints of two sides of a triangle is parallel to the third side and half the third side

⇒ EF || BC

⇒ EF || BD …(i)

And EF = × BC

But D is the midpoint of BC therefore 1/2× BC = BD

⇒ EF = BD …(ii)

E and D are midpoints of the sides AC and BC

⇒ ED || AB

⇒ ED || FB …(iii)

And ED = × AB

But F is the midpoint of AB therefore 1/2 × AB = FB

⇒ ED = FB …(iv)

Using (i), (ii), (iii) and (iv) we can say that BDEF is a parallelogram

Similarly we can prove that AFDE and FECD are also parallelograms

ii) as BDEF is parallelogram with FD as diagonal

the diagonal divides the area of parallelogram in two equal parts

⇒ area(ΔBFD) = area(ΔDEF) …(v)

as AFDE is parallelogram with FE as diagonal

the diagonal divides the area of parallelogram in two equal parts

⇒ area(ΔAFE) = area(ΔDEF) …(vi)

as CEFD is parallelogram with DE as diagonal

the diagonal divides the area of parallelogram in two equal parts

⇒ area(ΔEDC) = area(ΔDEF) …(vii)

From (v), (vi) and (vii)

area(ΔDEF) = area(ΔBFD) = area(ΔAFE) = area(ΔEDC) …(*)

from figure

⇒ area(ΔABC) = area(ΔDEF) + area(ΔBFD) + area(ΔAFE) + area(ΔEDC)

Using (*)

⇒ area(ΔABC) = area(ΔDEF) + area(ΔDEF) + area(ΔDEF) + area(ΔDEF)

⇒ area(ΔABC) = 4 × area(ΔDEF)

⇒ area(ΔDEF) = 1/4× area(ΔABC)

iii) from figure

⇒ area(ΔABC) = area(ΔDEF) + area(ΔBFD) + area(ΔAFE) + area(ΔEDC)

Using (*)

⇒ area(ΔABC) = area(ΔDEF) + area(ΔBFD) + area(ΔFED) + area(ΔEDC) …(viii)

From figure

area(ΔDEF) + area(ΔBFD) = area(BDEF) …(ix)

using (*)

area(ΔDEF) + area(ΔDEF) = area(BDEF)

area(ΔFED) + area(ΔEDC) = area(DCEF) …(x)

using (*)

area(ΔDEF) + area(ΔDEF) = area(DCEF)

therefore area(BDEF) = area(DCEF) …(xi)

substituting equation (ix), (x) and (xi) in equation (viii)

⇒ area(ΔABC) = area(BDEF) + area(BDEF)

⇒ area(ΔABC) = 2 × area(BDEF)

⇒ area(BDEF) = 1/2× area(ΔABC)