(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
But a > b, c > d b+d is not right.
That is to say, if A+C > B+D holds, A > B and C > D may not be obtained.
For example, a=2, b