Question

given $\triangle abc$ with $a^{2}+b^{2}=c^{2}$, write a paragraph proof of the converse of the pythagorean theorem.
given: $\triangle abc$, where $a^{2}+b^{2}=c^{2}$
prove: $\triangle abc$ is a right triangle.

draw perpendicular lines m and n intersecting at z. on line m, plot x so that $xz = b$. on line n, plot y so that $yz = a$. draw xy with length z.
by the pythagorean theorem, $a^{2}+b^{2}=\square$.

Explanation:

First, construct right triangle $XZY$ with perpendicular lines $m$ and $n$ intersecting at $Z$, $XZ = b$, $YZ = a$, and hypotenuse $XY = z$. By the Pythagorean Theorem, $a^2 + b^2 = z^2$. We are given that for $\triangle ABC$, $a^2 + b^2 = c^2$, so $z^2 = c^2$, meaning $z = c$. By the SSS (Side-Side-Side) Congruence Postulate, $\triangle ABC \cong \triangle XYZ$. Since $\triangle XYZ$ is a right triangle (by construction), $\triangle ABC$ must also be a right triangle.

For the blank: By applying the Pythagorean Theorem to the constructed right triangle $XZY$, the sum of the squares of the legs equals the square of the hypotenuse.

Answer:

Paragraph Proof:

Construct right triangle $XZY$ where lines $m$ and $n$ are perpendicular at $Z$, $XZ = b$, $YZ = a$, and hypotenuse $XY = z$. By the Pythagorean Theorem, $a^2 + b^2 = z^2$. Given $\triangle ABC$ has $a^2 + b^2 = c^2$, so $z^2 = c^2$, hence $z = c$. By SSS congruence, $\triangle ABC \cong \triangle XYZ$. Since $\triangle XYZ$ is right-angled, $\triangle ABC$ is a right triangle.

Blank Answer:

$z^2$