Question

In the figure, ABC adn BDC are two equilateral triangles such that D is the mid point of BC. Ae intersects BC in F.
$\left(i\right)ar\left(△BDE\right)=\frac{1}{4}ar\left(△ABC\right)\left(ii\right)ar\left(△BDE\right)=\frac{1}{2}ar\left(△BAE\right)\left(iii\right)ar\left(△BFE\right)=ar\left(△AFD\right)\left(iv\right)ar\left(△ABC\right)=2ar\left(△BEC\right)\left(v\right)ar\left(△BFE\right)=2ar\left(EFD\right)\left(vi\right)ar\left(△FED\right)=\frac{1}{8}ar\left(triangleAFC\right)$

Answer 1

Given : ABC and BDE are two equilateral triangles and D is mid point of BC. Ae intersects BC in F
To Prove
$\left(i\right)ar\left(△BDE\right)=\frac{1}{4}ar\left(△ABC\right)\left(ii\right)ar\left(△BDE\right)=\frac{1}{2}ar\left(△BAE\right)\left(iii\right)ar\left(△BFE\right)=ar\left(△AFD\right)\left(iv\right)ar\left(△ABC\right)=2ar\left(△BEC\right)\left(v\right)ar\left(△BFE\right)=2ar\left(EFD\right)\left(vi\right)ar\left(△FED\right)=\frac{1}{8}ar\left(triangleAFC\right)$
Proof: Let AB = BC = CA = x, then $BD=\frac{x}{2}$
(i) Now area of equilateral $△ABC=\frac{\sqrt{3}}{4}(side{)}^2=\frac{\sqrt{3}}{4}{x}^{2}c{m}^2$
and area of equilateral $△BED=\frac{\sqrt{3}}{4}{\left(\frac{x}{2}\right)}^2=\frac{\sqrt{3}}{4}\times \frac{x^2}{4}=\frac{1}{4}\left(\frac{\sqrt{3}}{4}{x}^2\right)$
$\frac{1}{4}$ (area of $△ABC)$
(ii) $\therefore △ABC=\angle BED$ are equilateral triangle
$\therefore \angle ACB=\angle DBE={60}^{\circ }$
But these are alternate angles
$\therefore AB||DEar\left(△BAE\right)=2ar\left(△BEC\right)=2ar\left(△BDE\right)\because EDisamedianof△EBC)\Rightarrow ar(△BDE)=\frac{1}{2}ar(△BAE)$
(iii) $\therefore △ABCand△BDE$ are equlateral triangles
$\therefore \angle ABC={60}^{\circ }and\angle BDE={60}^{\circ }\Rightarrow \angle ABC=\angle BDE$
But these are alternate angles
$\therefore AB||DE\therefore ar\left(△BED\right)=ar\left(△AED\right)Subtractingar\left(△EFD\right)frombothsidesar\left(△BED\right)-ar\left(\angle EFD\right)=ar\left(\angle AED\right)-ar\left(\angle EFD\right)\therefore ar\left(△BFE\right)=ar\left(△AFD\right)\left(iv\right)\because Edisamedianof\left(△BDE\right)\Rightarrow ar\left(△BEC\right)=2\times \frac{1}{4}ar\left(△ABC\right)\because ar\left(△BDE\right)=\frac{1}{2}ar\left(△ABC\right)=\frac{1}{2}ar\left(△ABC\right)$
(v) Let h be the height of vertex E, corresponding to the side BD in triangle BDE,
Let H be the height of vertex A, corresponding to the side BC in triangle ABC.
From part (i) ,
$ar\left(△BDE\right)=\frac{1}{4}ar\left(△ABC\right)\Rightarrow \frac{1}{2}\times BD\times h=\frac{1}{4}(\frac{1}{2}\times BC\times H)\Rightarrow BD\times =\frac{1}{4}(2BD\times H)\Rightarrow h=\frac{1}{2}HFromPart\left(iii\right)ar\left(△BFE\right)=ar\left(△AFD\right)=\frac{1}{2}\times FD\times H=\frac{1}{2}\times FD\times 2h=2(\frac{1}{2}\times FDtimesh)=2ar\left(△EFD\right)$
(vi) $ar\left(△AFC\right)=ar\left(△AFD\right)+ar\left(△ADC\right)=ar\left(△BFE\right)+\frac{1}{2}ar\left(△ABC\right)$
(Using part (iiii) and AD is the median of $\left(△ABC\right)$
$=ar\left(△BFE\right)+\frac{1}{2}\times 4ar\left(△BDE\right)=ar\left(△BFE\right)+2ar\left(△BDE\right)Now,frompart\left(v\right),ar\left(△BFE\right)=2ar\left(△FED\right)ar\left(△BDE\right)=ar\left(△BFE\right)+ar\left(△FED\right)=2ar\left(△FED\right)+ar\left(△FED\right)=3ar\left(△FED\right)$
from (ii), (iii) and (iv), we get
$ar\left(△AFC\right)=2ar\left(△FED\right)+2\times 3ar\left(△FED\right)=8ar\left(△FED\right)$