Kari wants to prove the Pythagorean theorem. To start her proof, she draws $ \triangle A B C $, where $ \overline{A B} \perp \overline{B C} $ and $ \overline{B D} \perp \overline{A C} $.
Complete the proof that $ A B^{2}+B C^{2}=A C^{2} $.
$ \overline{A B} \perp \overline{B C} $, and $ \overline{B D} \perp \overline{A C} $. Since perpendicular lines form right angles and all right angles are congruent, $ \angle A B C \cong \angle A D B $. Alsd, by the
So, $ \triangle A B C \sim \triangle A D B $ by the Similarity Theorem. Therefore $ \frac{A B}{A C}= $ You can rewrite this equation as $ A B^{2}=A C \cdot A D $.
Similarly, since $ \overline{A B} \perp \overline{B C} $ and $ \overline{B D} \perp \overline{A C}, \angle A B C \cong \angle B D C $. Also, by the Reflexive Property of Congruence, and so, by the Angle-Angle Similarity Theorem, $ \triangle A B C \sim $
₹. Therefore, $ \frac{A C}{B C}=\frac{B C}{D C^{\prime}} $, which you can rewrite as
Now, by the Addition Property of Equality and substitution, $ A B^{2}+B C^{2}=A C \cdot A D+A C \cdot D C $.
So, $ A B^{2}+B C^{2}= $
by the Distributive Property. Since $ A D+D C= $
AC by the
, it must be true that $ A B^{2}+B C^{2}= $
$ A C \cdot A C $, or $ A B^{2}+B C^{2}=A C^{2} $.

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