Question

ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC such that CE = 2 DE and F is the point of BC such that BF = 2 FC. Prove that:
$\left(i\right)ar\left(ADEG\right)=ar\left(GBCE\right)\left(ii\right)ar\left(△EGB\right)-\frac{1}{6}ar\left(ABCD\right)\left(iii\right)ar\left(△EFC\right)=\frac{1}{2}ar\left(△EBF\right)\left(iv\right)ar\left(△EGB\right)=\frac{3}{2}ar\left(△EFC\right)$
(v) Find what portion of the ara of parallelogram is the area of $△EFG.$

Answer 1

Construction : Draw $EP\perp ABandEQ\perp BC$

Proof : We have, AG = 2GB and CE = 2DE and BF = 2FC
$\Rightarrow AB-GB=2GBandCD-DE=2DEandBC-FC=2FC\Rightarrow AB=3BGandCD=3DEandBC=3FC\Rightarrow GB=\frac{1}{3}ABandDE=\frac{1}{3}CDandFC=\frac{1}{3}BC$
(i)$ar\left(ADEG\right)=\frac{1}{2}(AG+6DE)\times EP\Rightarrow ar\left(ADEG\right)=\frac{1}{2}(\frac{2}{3}AB+\frac{1}{3}CD)\times EP\Rightarrow ar\left(ADEG\right)=\frac{1}{2}(\frac{2}{3}AB+\frac{1}{3}B)\times EP\Rightarrow ar\left(ADEG\right)=\frac{1}{2}\times AB\times EPandar\left(GBCE\right)=\frac{1}{2}(GB+CE)\times EP\Rightarrow ar\left(GBCE\right)=\frac{1}{2}(\frac{1}{3}AB+\frac{2}{3}CD)\times EP\Rightarrow ar\left(GBCE\right)=\frac{1}{2}[\frac{1}{3}AB+\frac{2}{3}AB]\times EP\Rightarrow ar\left(GBCE\right)=\frac{1}{2}\times AB\times EP$
Compare equation (ii) and (iii) ar (ADEG) = ar(GBCE)
$\left(ii\right)ar\left(△EGB\right)=\frac{1}{2}\times GB\times EP\Rightarrow ar\left(△EGB\right)=\frac{1}{2}\times \frac{1}{3}AB\times EP=\frac{1}{6}AB\times EP=\frac{1}{6}ar\left(|{|}^{gm}ABCD\right)$
(iii)$ar\left(△EFC\right)=\frac{1}{2}\times FC\times EQandar\left(△EBF\right)=\frac{1}{2}\times BF\times EQ\Rightarrow ar\left(△EBF\right)=\frac{1}{2}\times BF\times EQ\Rightarrow ar\left(△EBF\right)=\frac{1}{2}\times 2FC\times EQ$
Compare equation (iv) and (v)
$ar\left(△EFC\right)=\frac{1}{2}\times ar\left(△EBF\right)\left(iv\right)from\left(ii\right)partar\left(△EGB\right)=\frac{1}{6}ar\left(|{|}^{gm}ABCD\right)From\left(iii\right)Partar\left(△EFC\right)=\frac{1}{2}ar\left(△EBF\right)\Rightarrow ar\left(△EFC\right)=\frac{1}{2}ar\left(△EBF\right)\Rightarrow ar\left(△EFC\right)=\frac{1}{3}ar\left(△EBC\right)\Rightarrow ar\left(△EFC\right)=\frac{1}{3}\times \frac{1}{2}CE\times EP=\frac{1}{3}\times \frac{1}{2}\times \frac{2}{2}D\times EP=\frac{1}{6}\times \frac{2}{3}\times ar\left(|{|}^{gm}ABCD\right)\Rightarrow ar\left(△EFC\right)=\frac{2}{3}\times ar\left(△EGB\right)\Rightarrow ar\left(△EGB\right)=\frac{3}{2}ar\left(△EFC\right)\left(v\right)ar\left(△EFG\right)=ar(trap.BGEC)-ar\left(△BGF\right)$
Now, ar (trap. BGEC) $=\frac{1}{2}(GB+EC)\times EP$
$=\frac{1}{2}(\frac{1}{3}AB+\frac{2}{3}CD)\times EP=\frac{1}{2}AB\times EP=\frac{1}{2}ar\left(|{|}^{gm}ABCD\right)ar\left(△EFC\right)=\frac{1}{9}ar\left(|{|}^{gm}ABCD\right)$
And $ar\left(△BGF\right)=\frac{1}{2}BF\times GR$
$\frac{1}{2}\times \frac{2}{3}BC\times GR=\frac{2}{3}\times \frac{1}{2}BC\times GR=\frac{2}{3}\times ar\left(△GBC\right)=\frac{2}{3}\times \frac{1}{2}GB\times EP=\frac{1}{3}\times \frac{1}{3}AB\times EP=\frac{1}{9}AB\times EP=\frac{1}{9}ar\left(|{|}^{gm}ABCD\right)$
$\therefore$ From (vii)
$ar\left(△EFG\right)=\frac{1}{2}ar\left(|{|}^{gm}ABCD\right)-\frac{1}{9}ar\left(|{|}^{gm}ABCD\right)-\frac{1}{9}ar\left(|{|}^{gm}ABCD\right)=\frac{5}{18}ar\left(|{|}^{gm}ABCD\right)$