Question

sᴏʟᴠᴇ

❌ɴᴏ ꜱᴩᴀᴍᴍɪɴɢ ❌​

Answer 1

Given in ΔABC, D, E and F are midpoints of sides AB, BC and CA respectively.

BC = EC

Recall that the line joining the midpoints of two sides of a triangle is parallel to third side and half of it.

Hence DF = (1/2) BC

⇒ (DF/BC) = (1/2) → (1)

Similarly, (DE/AC) = (1/2) → (2)

(EF/AB) = (1/2) → (3)

From (1), (2) and (3) we have

But if in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar

Hence ΔABC ~ ΔEDF [By SSS similarity theorem]

Area of def/abc = DF^2/bc^2

Hence area of ΔDEF : area of ΔABC = 1 : 4

Answer 2

$\huge\underline\mathcal\blue {Solution}$

Given that in triangle ABC , D , E and F are mid points of AB , BC and AC respectively .

Points D and E are mid points of AB and AC respectively .

So , using mid - point theorem ,

$DF= \frac{1}{2} BC$

and DF || BC .

DF = BF and DF || BF

Similarly , EF = BD and EF || BD .

=> BEFD is a ||gm .

so , triangle BDE is congruent to triangle FED . ....(i)

(diagonals divides a ||gm into two congruent triangles .)

Similarly, ECFD and ADEF are also parallelogram .

=> Traingle ECF is congruent to triangle FDE . .....(ii)

and triangle FED is congruent to triangle DAF . .....(iii)

From (i) , (ii) and (iii)

=> Triangle BDE is congruent to triangle FED is congruent to triangle EFC is congruent to triangle DAF .

ar ( BDE ) = ar ( FED )

= ar ( EFC ) = ar ( DAF )

(conguent triangles have same area)

Also , in triangles ABC and EFD ,

<A = <E

<B = <F

=> triangle ABC ~ Triangle EFD ( By AA similarity )

Hence ,

ar(ABC )/ar(DEF)=4/1

= 4:1