如图,△ABC中,∠BAC=90°,AD⊥BC于点D,△ABE和△ACF都是等边三角形,若AD:BC=12:25,且AB>AC,求:S△DBES△DAF.

问题描述:

如图,△ABC中,∠BAC=90°,AD⊥BC于点D,△ABE和△ACF都是等边三角形,若AD:BC=12:25,且AB>AC,求:

S△DBE
S△DAF

∵∠BAC=90°,AD⊥BC,∴∠ABD+∠BAD=∠CAD+∠BAD=90°∴∠ABD=∠CAD ①∵△ABE和△ACF都是正三角形,∴∠ABE=∠CAF ②①+②,得∠DBE=∠DAF ③∵AD⊥BC,∴∠BDA=∠ADC=90° ④∴由 ①和 ④可知△ABD∽△CAD∴BDAD...