Question Number 171717 by infinityaction last updated on 20/Jun/22
Answered by mr W last updated on 20/Jun/22
Commented by mr W last updated on 21/Jun/22
![let AH=AE ⇒ΔAED≡ΔAHD ⇒ED=HD ΔEGB≡ΔADC ⇒EG=AD ⇒DG//AE, DG=AE ⇒[ΔEGD]=[ΔAED]=[ΔAHD] ∠DFG=∠FDE=γ ⇒∠CFD=∠CDF=∠ADE=∠ADH ⇒HD//CF [ΔDFG]=[ΔDFE]=[ΔDFH]=[ΔDCH] [EGFD]=[ΔEGD]+[ΔDFG] =[ΔAHD]+[ΔDCH] =[ΔADC] blue area=[ΔEGB]+[EGFD] =[ΔADC]+[ΔADC] =2×[ΔADC] =2×yellow area ✓](https://www.tinkutara.com/question/Q171769.png)
$${let}\:{AH}={AE} \\ $$$$\Rightarrow\Delta{AED}\equiv\Delta{AHD} \\ $$$$\Rightarrow{ED}={HD} \\ $$$$ \\ $$$$\Delta{EGB}\equiv\Delta{ADC} \\ $$$$\Rightarrow{EG}={AD} \\ $$$$\Rightarrow{DG}//{AE},\:{DG}={AE} \\ $$$$\Rightarrow\left[\Delta{EGD}\right]=\left[\Delta{AED}\right]=\left[\Delta{AHD}\right] \\ $$$$ \\ $$$$\angle{DFG}=\angle{FDE}=\gamma \\ $$$$\Rightarrow\angle{CFD}=\angle{CDF}=\angle{ADE}=\angle{ADH} \\ $$$$\Rightarrow{HD}//{CF} \\ $$$$ \\ $$$$\left[\Delta{DFG}\right]=\left[\Delta{DFE}\right]=\left[\Delta{DFH}\right]=\left[\Delta{DCH}\right] \\ $$$$ \\ $$$$\left[{EGFD}\right]=\left[\Delta{EGD}\right]+\left[\Delta{DFG}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[\Delta{AHD}\right]+\left[\Delta{DCH}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[\Delta{ADC}\right] \\ $$$$ \\ $$$${blue}\:{area}=\left[\Delta{EGB}\right]+\left[{EGFD}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[\Delta{ADC}\right]+\left[\Delta{ADC}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}×\left[\Delta{ADC}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}×{yellow}\:{area}\:\checkmark \\ $$
Commented by infinityaction last updated on 21/Jun/22
$${thank}\:{you}\:{sir} \\ $$
Commented by Tawa11 last updated on 25/Jun/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$