Question

如图,在△ABC中,CD⊥AB与点D,过点D作点DE//BC交AC与点E.若AB=14,AD/BD=5/9,CD=12.

Answer 1

SΔABC=1/2AB×CD=84,

∵DE∥BC,

∴ΔADE∽ΔABC,

∴SΔADE/SΔABC=(AD/AB)^2=(5/14)^2=25/196

∴SΔADE=75/7.

∵AC=√(AD^2+CD^2)=13,

BC=√(BD^2+CD^2)=15,

又DE/BC=AD/AB,DE/15=5/14,

DE=75/14,

AE/AC=5/14,

∴AE=65/14,

∴ΔADE周长=5+75/14+65/14=15.