1, at △BCD and △BCA,
∠B=∠B,∠BDC=∠ACB
So △BCD∽△BCA
∴AB/BC=BC/BD
That is BD=BC? /AB= 12? / 13= 144/ 13
Then ad =13-(144/13) = 25/13.
∴CD? =(25/ 13)×( 144/ 13)
That is CD=60/ 13.
∴bd= 144/ 13,cd=60/ 13
2、3? =BD×AD ①
5? =BD×(BD+AD)? ②
The solution is BD=4 and CD=9/4.
∴AB=4+(9/4)=25/4
AC=√[(9/4)×(25/4)]= 15/4
3、AD=AC? /BD=4/3
CD=√(AD×BD)=√[(4/3)×3]=2
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