& lt5> connects BC, because CD is perpendicular to AB, which is the midpoint of AB, and triangle BAC is an isosceles triangle (three lines in one), so AC=BC.

It is also concluded that AB=BC, so AC=AB (equivalent substitution).

& lt 1 1 & gt； Because AD squared angle BAC, DE vertical AB and DF vertical AC, DE=DF. AE=AF because ∠EAD=∠FAD, ∠AED=∠AFD and DE=DF, △AED=△AFD. So AD divides EF vertically.

& lt 14 & gt； Because CD=CE, then ∠CDE=∠CED. Because ∠ACB=∠CDE+∠CED, ∠ACB= 1/2∠E, ∠DBC=∠CED, DB=DE.