Question Number 60588 by behi83417@gmail.com last updated on 22/May/19
Commented by behi83417@gmail.com last updated on 22/May/19
$${AD}\:\&{CE}\:{are}\:{angular}\:{bisector}\:{of} \\ $$$$\measuredangle{A}\:\&\measuredangle{C}. \\ $$$${DG}\bot{AB},{EF}\bot{BC},{BH}\bot{DE},{DE}=\frac{\mathrm{1}}{\mathrm{2}}{AC}. \\ $$$$\Rightarrow\frac{{BH}}{{DG}+{EF}}=? \\ $$