Zoe is using the figure shown below to prove Pythagorean Theorem using triangle similarity:

In the given triangle ABC, angle A is 90o 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?

Question

Zoe is using the figure shown below to prove Pythagorean Theorem using triangle similarity:

In the given triangle ABC, angle A is 90o 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?

anonymous · Answer

attachment

anonymous · Answer

By distribution, AC2 plus AB2 = BC multiplied by the quantity DC plus AB.
 By distribution, AC2 plus AB2 = BC multiplied by the quantity DC plus BD.
 By distribution, AC2 plus AD2 = BC multiplied by the quantity DC plus AB.
 By distribution, AC2 plus AD2 = BC multiplied by the quantity DC plus BD.

anonymous · Answer

i cant help you but can you see if you can help me real quick

anonymous · Answer

you cant help me? @Lil_Noahz

anonymous · Answer

i dont know maybe i have to look

anonymous · Answer

Ok so what are you struggling with in the question?

anonymous · Answer

@phi can you help me please? i think its the third one but im not sure

anonymous · Answer

i offered help what is it that you need help with./?

anonymous · Answer

to check my answer @Lil_Noahz

anonymous · Answer

ok

anonymous · Answer

it might be imnot sure

phi · Answer

I would use the idea that the segments BD+DC add up to the whole side BC

anonymous · Answer

i dont want ot say yes and it wrong because that happend to me alot on this

phi · Answer

the whole line segment  BC=  BD + DC
then if you multiply both sides by BC you get
$$ BC \cdot BC =BC(BD+DC) \BC^2=BC(BD+DC)  $$

phi · Answer

that is a good start.  now distribute the BC on the right side
$$ BC^2 = BC \cdot BD + BC\cdot DC $$

anonymous · Answer

ok im writing it down then what?

phi · Answer

you are trying to prove pythagoras,  which for this triangle means
$$ BC^2 = AB^2 + AC^2 $$
we know (so far)
$$ BC^2 = BC \cdot BD + BC\cdot DC $$
so we have to show BC*BD is AB*AB 
and BC*DC is AC*AC

which we can do using similar triangles.

but you have enough of a hint.
which of your choices shows BC*BD +BC*DC ?

phi · Answer

or , the equivalent  BC*(BD+DC) ?

anonymous · Answer

the second one? @phi

phi · Answer

yes. the 2nd choice is the only one that is on the "right track"