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Helppppp 15P!!
Brainliest if correct!!

Use the drop-down menus to complete the proof of the Pythagorean theorem.

By the angle-angle similarity postulate, △YXZ∼△YZQ and △YXZ∼△ZXQ. Since similar triangles have
sides, ac=fa and bc=eb . Solving the equation for a² and b² gives a2=cf and b2= . Adding these together gives a2+b2=cf+ce. Factoring out the common segment gives a2+b2=c(f+e). Using gives a2+b2=c(c), which simplifies to a2+b2=c2 .

Helppppp 15P!! Brainliest if correct!! Use the drop-down menus to complete the proof of the Pythagorean theorem. By the angle-angle similarity postulate, △YXZ∼△ class=

Answer :

milanc6525

Answer:

First Drop: Proportional

Second Drop: cf

Third Drop: [I think] Segment Addition Postulate

Step-by-step explanation:

I just looked through several websites with the same question. I have come to a conclusion that these are perhaps correct. (I'm taking the test right now)

Answer:

1. Proportional.

2. ce

3. Segment Addition Postulate.

Step-by-step explanation:

By the angle-angle similarity postulate, △YXZ∼△YZQ and △YXZ∼△ZXQ.

The corresponding sides of two similar triangles are proportional.

Since similar triangles have  proportional sides, therefore

[tex]\frac{a}{c}=\frac{f}{a}[/tex] and [tex]\frac{b}{c}=\frac{e}{b}[/tex]

Solving the equation for a² and b² gives

[tex]a^2=cf[/tex] and [tex]b^2=ce[/tex]

The value of a² is cf and the value b² is ce.

Adding these together gives

[tex]a^2+b^2=cf+ce[/tex]

Factoring out the common segment gives

[tex]a^2+b^2=c(f+e)[/tex]

From the given figure it is clear that

[tex]c=f+e[/tex]         (Segment Addition Postulate)

Using segment Addition Postulate, we get

[tex]a^2+b^2=c(c)[/tex]

On simplification, we get

[tex]a^2+b^2=c^2[/tex]

Therefore the required answers are 1. Proportional, 2. ce, 3. Segment Addition Postulate.

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