Question

Zoe is using the figure shown below to prove Pythagorean Theorem using triangle similarity:
In the given triangle ABC, angle A is 90 degrees and segment AD is perpendicular to segment BC.

Answer 1

Which of these could be a step to prove that BC^2 = AB^2 + AC^2?

By distribution, AC^2 plus AB^2 = BC multiplied by the quantity DC plus AB.
By distribution, AC^2 plus AB^2 = BC multiplied by the quantity DC plus BD.
By distribution, AC^2 plus AD^2 = BC multiplied by the quantity DC plus AB.
By distribution, AC^2 plus AD^2 = BC multiplied by the quantity DC plus BD.

Answer 2

@jim_thompson5910

Answer 3

Could you help me with this? Its the same question I have

Answer 4

@ParthKohli HI, can you please help me with this?

Answer 5

@phi help me with this, please?

Answer 6

the question is pretty nebulous. But it looks like they are using the idea that
segment BC is the same as BD+DC
so BC^2 = BC * BC = BC* (BD+DC)

Answer 7

so that narrows it down to answers D and B, right?

Answer 8

also, it seems they are using
AB^2 + AC^2 = BC^2

Answer 9

so we have
AB^2 + AC^2 = BC* (BD+DC)

I'm not sure how this is a proof, but it's true.

Answer 10

and the other choices don't look true.

Answer 11

That makes much more sense, you're a life saver thank you so much!