(2) verification: AD? =BD×CD

(3) Can I get AB? =BD×BC

(4) Can I get AC? =CD×BC

Prove:

(2)

∵AD⊥BC

∴∠ADB=∠ADC=90

∠∠BAD+∠DAC =∠BAC = 90

∠C+∠DAC= 180 -∠ADC=90

∴∠BAD=∠C

∴△ABD∽△ACD

∴BD/AD=AD/CD

Namely: AD? =BD×CD

(3)

∵AD⊥BC

∴∠ADB=90 =∠BAC

∫∠BAD+∠B = 180-∠ADB = 90

∠C+∠B= 180 -∠BAC=90

∴∠BAD=∠C

∴△ABD∽△ABC

∴AB/BC=BD/AB

Namely: AB? =BD×BC

(3)

∵AD⊥BC

∴∠ADC=90 =∠BAC

∠∠DAC+∠C = 180-∠ADC = 90

∠B+∠C= 180 -∠BAC=90

∴∠DAC=∠B

∴△ACD∽△ABC

∴AC/BC=CD/AC

Namely: AC? =CD×BC