Correct option is (C) BC . DC
In right \(\triangle BAD,\)
\(\angle A=90^\circ\) and
\(\angle A+\angle B+\angle D=180^\circ\)
\(\Rightarrow\) \(\angle B+\angle D=90^\circ\) ______________(1) \((\because\angle A=90^\circ)\)
In right \(\triangle ACB,\)
\(\angle C=90^\circ\) \((\because\angle AC\bot BD)\)
\(\therefore\) \(\angle B+\angle 1=90^\circ\) ______________(2)
From (1) & (2), we obtain
\(\angle D=\angle 1\)
Now in triangles \(\triangle ACB\;\&\;\triangle DCA,\)
\(\angle 1=\angle D\)
\(\angle ACB=\angle DCA=90^\circ\) \((\because AC\bot BD)\)
\(\therefore\) \(\triangle ACB\sim\triangle DCA\) (By AA similarity rule)
\(\therefore\) \(\frac{AC}{CD}=\frac{BC}{AC}\)
\(\Rightarrow AC^2=BC.DC\)
