Question

Hillary is using the figure shown below to prove Pythagorean Theorem using triangle similarity: In the given triangle ABC, angle A is 90° and segment AD is perpendicular to segment BC. The figure shows triangle ABC with right angle at A and segment AD. Point D is on side BC. Which of these could be a step to prove that BC2 = AB2 + AC2? (6 points) By the addition property of equality, AC2 plus AD2 = AB multiplied by DC plus AD2. By the addition property of equality, AC2 plus AD2 = BC multiplied by DC plus AD2. By the addition property of equality, AC2 plus AB2 = AB multiplied by DC plus AB2. By the addition property of equality, AC2 plus AB2 = BC multiplied by DC plus AB2.

Answer 1

\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}