Converse Pythagorean Theorem

Consider a triangle $A B C$ with side lengths $a,$ $b,$ and $c$ such that $c^{2} = a^{2} + b^{2} .$ The idea is to prove that $m ∠ C = 9 0^{\circ} .$

Now, construct a right triangle

P Q R

such that

P R = b,

Q R = a,

and

m ∠ R = 9 0^{\circ} .

Write

P Q

in terms of

a

and

b

P Q^{2} = a^{2} + b^{2}

Remember that

c^{2} = a^{2} + b^{2} .

Then, set the left-hand sides of these equations equal to each other.

c^{2} = P Q^{2} \Rightarrow c^{2} c = P Q^{2} = P Q

The last equation implies that the sides of

△ A B C

are congruent to the sides of

△ P Q R .

a b c = Q R \Rightarrow = P R \Rightarrow = P Q \Rightarrow \overline{B C} \overline{A C} \overline{A B} ≅ \overline{Q R} ≅ \overline{P R} ≅ \overline{P Q}

Triangles

A B C

and

P Q R

are congruent by the Side-Side-Side Congruence Theorem. This implies that

∠ C ≅ ∠ R,

because corresponding angles are congruent. Remember, by construction

m ∠ R = 9 0^{\circ} .

Therefore,

m ∠ C = 9 0^{\circ}

which completes the proof.

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