Question

angle - angle similarity theorem. therefore (\frac{}{ac}=\frac{}{ab}). you can rewrite this equation as (ab = accdot ad).
similarly, since (overline{ab}perpoverline{bc}) and (overline{bd}perpoverline{ac}), (angle abccongangle bdc). also, by the reflexive property of congruence, and so, by the angle - angle similarity theorem, (\triangle abcsim). therefore, (\frac{ac}{bc}=\frac{bc}{dc}), which you can rewrite as.
now, by the addition property of equality and substitution, (ab^{2}+bc^{2}=accdot ad + accdot dc). so, (ab^{2}+bc^{2}) by the distributive property. since (ad + dc = ac) by the, it must be true that (ab^{2}+bc^{2}=accdot ac), or (ab^{2}+bc^{2}=ac^{2}).

Explanation:

Step1: Identify similar angles and triangles

Since \(AB\perp BC\) and \(BD\perp AC\), \(\angle ABC\cong\angle BDC\). Also, \(\angle C\cong\angle C\) by the Reflexive Property of Congruence. By the Angle - Angle Similarity Theorem, \(\triangle ABC\sim\triangle BDC\).

Step2: Use the similarity of triangles to get the proportion

For similar triangles \(\triangle ABC\) and \(\triangle BDC\), we have \(\frac{AC}{BC}=\frac{BC}{DC}\), which can be rewritten as \(BC^{2}=AC\cdot DC\).

Step3: Apply the distributive property

We know that \(AB^{2}+BC^{2}=AC\cdot AD + AC\cdot DC\). By the Distributive Property \(AC\cdot AD+AC\cdot DC = AC(AD + DC)\). Since \(AD + DC=AC\) by the Segment Addition Postulate.

Answer:

\(\angle C\cong\angle C\); \(\triangle BDC\); \(BC^{2}=AC\cdot DC\); \(AC(AD + DC)\); Segment Addition Postulate