Q

In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that (ii) ar (BDE) = 1/2 ar (BAE)

Q: 5    In Fig. $\small 9.33$, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that

(ii)  $\small ar(BDE)=\frac{1}{2}ar(BAE)$

[Hint: Join EC and AD. Show that $\small BE\parallel AC$ and  $\small DE\parallel AB$, etc.]

Views

Since $\Delta$ABC and $\Delta$ BDE are equilateral triangles.
Therefore, $\angle$ACB = $\angle$DBE = $60^0$
$\Rightarrow$ BE || AC

$\Delta$BAE and $\Delta$BEC are on the same base BE and between same parallels BE and AC.
Therefore, ar ($\Delta$BAE) = ar($\Delta$BEC)
$\Rightarrow$ ar($\Delta$BAE) = 2 ar($\Delta$BED)   [since D is the meian of $\Delta$BEC ]
$\Rightarrow$ $\small ar(BDE)=\frac{1}{2}ar(BAE)$
Hence proved.

Exams
Articles
Questions