Question

1. given △abc with (a^2 + b^2 = c^2), write a paragraph proof of the converse of the pythagorean theorem. see example 2
2. write a two - column proof of the (45^{circ}-45^{circ}-90^{circ}) triangle theorem. see example 3

Explanation:

Problem 17:

We start by constructing a right triangle $\triangle DEF$ with right angle at $F$, where $DE = c$, $EF = a$, $DF = b$. By the Pythagorean Theorem, $a^2 + b^2 = DE^2$. Since we know $a^2 + b^2 = c^2$ for $\triangle ABC$, $DE = c = AB$. By the SSS Congruence Postulate, $\triangle ABC \cong \triangle DEF$. Corresponding angles of congruent triangles are equal, so $\angle C = \angle F = 90^\circ$, proving $\triangle ABC$ is right-angled.

We use the given isosceles right triangle, define leg length $x$, apply the Pythagorean Theorem to find the hypotenuse, then summarize the relationship between legs and hypotenuse for a $45^\circ$-$45^\circ$-$90^\circ$ triangle.

Answer:

Paragraph Proof:
Let $\triangle ABC$ be a triangle with side lengths $a$, $b$, $c$ such that $a^2 + b^2 = c^2$. Construct a right triangle $\triangle DEF$ with a right angle at $F$, where $EF = a$, $DF = b$, and hypotenuse $DE$. By the Pythagorean Theorem, $EF^2 + DF^2 = DE^2$, so substituting the given side lengths gives $a^2 + b^2 = DE^2$. Since $a^2 + b^2 = c^2$, we can conclude $DE^2 = c^2$, so $DE = c$. Now, $\triangle ABC$ and $\triangle DEF$ have all corresponding sides equal: $AB = DE = c$, $BC = EF = a$, and $AC = DF = b$. By the Side-Side-Side (SSS) Triangle Congruence Postulate, $\triangle ABC \cong \triangle DEF$. Because corresponding angles of congruent triangles are congruent, $\angle C \cong \angle F$. Since $\angle F$ is a right angle, $\angle C$ must also be a right angle. Therefore, $\triangle ABC$ is a right triangle with a right angle at $C$, which proves the Converse of the Pythagorean Theorem.

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Problem 18: