Question

三角形ABC中，角BAC=45度，AD垂直BC于D,BD=3,DC=2.求三角形ABC的面积

Answer 1

解：作BE⊥AC于点E，交AD于点F

∵∠蔽谈BAC=45°

∴BE=AE

∵∠CBE+∠C=∠EAF+∠C=90°

∴△BCE≌△AFE

∵∠AEF=∠BEC=90°

∴AF=BD=3+2=5

设DF=x

∵∠CBE=∠CAD

∴△BDF∽宏悄碰△运铅ADC

∴DF/CD=BD/AD

即x/2=3/(5+x)

解得x=1

∴AD=5+1=6

∴S△ABC=1/2*5*6=15 

Answer 2

AD=BD/tanBAD = DC/巧空tan(45-BAD)
tan BAD/tan(45-BAD) = BD/DC = 1.5
2tan BAD= 3 (tan45 - tan BAD)/(1+tan45 tanBAD)
2tanBAD = 3(1-tanBAD)/(1+tanBAD)
tanBAD =1/州余2
AD = 2BD = 6
面积=1/2*AD*BC = 1/册宽滚2 * 6 * 5 = 15