Question

在三角形ABC中,D为BC边上的一点,BC=3BD,AD=根号2,角ADB=135°,若AC=根号2倍AB,求BD

Answer 1

作AH⊥BC于H,则AH=AD×SIN(180°-135°)=1=DH.则BH=BD+DH=BD+1,CH=2BD-1 又AB^2-AH^2=AB^2-1=BH^2=(BD+1)^2 ,则AB^2=(BD+1)^2+1 ; 又 AC^2-AH^2=2AB^2-AH^2=2AB^2-1=CH^2=(2BD-1)^2,则2AB^2=(2BD-1)^2+1; 所以,2(BD ...