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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