Question

select all the correct answers.
in \\( \triangle a b c, \overline { b d } \\) is perpendicular to \\( \overline { a c } \\) as shown in the figure.

which equations are true?
\\( \square b c ^ { 2 } + c d ^ { 2 } = b d ^ { 2 } \\)
\\( \square a c ^ { 2 } + c b ^ { 2 } = a b ^ { 2 } \\)
\\( \square a b ^ { 2 } + b c ^ { 2 } = a c ^ { 2 } \\)
\\( \square a b ^ { 2 } + b d ^ { 2 } = a d ^ { 2 } \\)
\\( \square a d ^ { 2 } + d b ^ { 2 } = a b ^ { 2 } \\)
\\( \square b d ^ { 2 } + d c ^ { 2 } = b c ^ { 2 } \\)

Explanation:

Step1: Apply Pythagorean theorem to \(\triangle ABD\)

In right - triangle \(\triangle ABD\) (right - angled at \(D\)), by the Pythagorean theorem \(AD^{2}+DB^{2}=AB^{2}\).

Step2: Apply Pythagorean theorem to \(\triangle BCD\)

In right - triangle \(\triangle BCD\) (right - angled at \(D\)), by the Pythagorean theorem \(BD^{2}+DC^{2}=BC^{2}\).

Step3: Apply Pythagorean theorem to \(\triangle ABC\)

In right - triangle \(\triangle ABC\) (right - angled at \(B\)), by the Pythagorean theorem \(AB^{2}+BC^{2}=AC^{2}\).

Answer:

\(\boxed{AB^{2}+BC^{2}=AC^{2}}\), \(\boxed{AD^{2}+DB^{2}=AB^{2}}\), \(\boxed{BD^{2}+DC^{2}=BC^{2}}\)