Respuesta :
\Delta BAC\sim \Delta ADC\sim \Delta BDA
from these we can establish the relations
AB^{2}=BD\times BC, AC^{2}=DC\times BC and AD^{2}=BD\times DC
consider AC^{2}=DC\times BC \Rightarrow AC^{2}+AD^{2}=DC\times BC+AD^{2}\rightarrow(B) is true
but it is not useful to prove the given theorem
again consider AC^{2}=DC\times BC \Rightarrow AC^{2}+AB^{2}=DC\times BC+AB^{2}\rightarrow(D) is true
\Rightarrow AC^{2}+AB^{2}=DC\times BC+BD\times BC
\Rightarrow AC^{2}+AB^{2}=BC(DC+BD)
\Rightarrow AC^{2}+AB^{2}=BC(BC)
\Rightarrow AC^{2}+AB^{2}=BC^{2}
Answer:
By the addition property of equality, AC2 plus AB2 = BC multiplied by DC plus AB2
Step-by-step explanation:
I took the test and got it right :)
