Question

Given: ΔABC is a right triangle. Prove: a2 + b2 = c2 Right triangle BCA with sides of length a, b, and c. Perpendicular CD forms right triangles BDC and CDA. CD measures h units, BD measures y units, DA measures x units. The following two-column proof with missing justifications proves the Pythagorean Theorem using similar triangles: Statement Justification Draw an altitude from point C to Line segment AB Let segment BC = a segment CA = b segment AB = c segment CD = h segment DB = x segment AD = y y + x = c c over a equals a over y and c over b equals b over x a2 = cy; b2 = cx a2 + b2 = cy + b2 a2 + b2 = cy + cx a2 + b2 = c(y + x) a2 + b2 = c(c) a2 + b2 = c2 Which is not a justification for the proof? Pieces of Right Triangles Similarity Theorem Side-Side-Side Similarity Theorem Substitution Addition Property of Equality

Answer 1

Hello, fellow Brainlies! The answer the the question that you have is B:

Side-Side-Side Similarity Theorem

I had that answer on my test and got it correct. Proof is in the image.

Hope I helped you get a 100%, and PLEASE mark me Brainliest!

Your fellow Brainlie,

~ AGirlGenius

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Answer 2

Answer:

There is a misprint in the question.

In the statement you have written DB=x ,and DA=y but in Question you have written DA= x and DB=y.

So, let me just considering your Statement justification

Let segment BC = a, segment CA = b ,segment AB = c segment CD = h, segment DB = x, segment AD = y ,y + x = c

In Δ B DC and Δ BC A

∠B D C =∠B C A [each being 90°]

∠ B is common.

Δ B D C is similar to Δ BC A.

$\frac{BD}{BC}=\frac{BC}{BA}\\\frac{x}{a}=\frac{a}{x}$

⇒ a² = c x  .........(1)

Similarly we can prove that

Δ ADC is similar to Δ BC A.

⇒b²= c y ......(2)

adding (1) and (2)

⇒$a^2+ b^2=cx+cy$

$a^2+ b^2=c(x+y)$

               =c×c [∴ x+y =c]

                 =c²

So, we have used two properties 1.  Right Triangles Similarity Theorem 2.Substitution Addition Property of Equality.

we haven't used  Side-Side-Side Similarity Theorem. It is not the right justification for the proof.