Question Number 47788 by behi83417@gmail.com last updated on 14/Nov/18
Commented by behi83417@gmail.com last updated on 14/Nov/18
$${ABCD}:{square}. \\ $$$$\measuredangle{AEC}=\theta,\measuredangle{BED}=\varphi. \\ $$$${show}\:{that}:\:\:\:\:\frac{{S}_{{B}\overset{\blacktriangle} {{E}D}} }{{S}_{{A}\overset{\blacktriangle} {{E}C}} }=\frac{{tg}\varphi}{{tg}\theta} \\ $$