Correct option is (A) ∠A = ∠B
Given that \(AD\bot BC\) \(and\;\angle A=90^\circ\)
In triangle ABC, we have
\(\angle A+\angle B+\angle C=180^\circ\) (Sum of angles in a triangle)
\(\Rightarrow\) \(\angle B+\angle C=180^\circ-\angle A\)
\(=180^\circ-90^\circ=90^\circ\)
\(\because\) \(\angle C\neq0^\circ\) \((\because\,If\,\angle C=0^\circ\,then\,A,B\;\&\;C\) do not form a triangle)
i.e., \(\angle C>0^\circ\)
\(\therefore\) \(\angle B<90^\circ\) \((\because\angle B+\angle C=90^\circ\;\&\;\angle C=0^\circ)\)
\(\therefore\) \(\angle B<\angle A\)
\(\Rightarrow\) \(\angle B\neq\angle A\)
Hence, option (A) is incorrect.
Now in triangles \(\triangle ABC\;\&\;\triangle DBA,\)
\(\angle ABC=\angle DBA\) (Common angles)
\(\angle BAC=\angle BDA=90^\circ\)
\(\therefore\) \(\triangle ABC\sim\triangle DBA\) ______________(1) (By AA similarity rule)
Now in triangles \(\triangle ABC\;and\;\triangle DAC,\)
\(\angle ACB=\angle DCA\) (Common angles)
\(\angle BAC=\angle ADC=90^\circ\)
\(\therefore\) \(\triangle ABC\sim\triangle DAC\) ______________(2) (By AA similarity rule)
From (1) & (2), we have
\(\triangle ABC\sim\triangle DBA\) \(and\;\triangle ABC\sim\triangle DAC\)
\(\Rightarrow\) \(\triangle DBA\sim\triangle ABC\sim\triangle DAC\)
\(\Rightarrow\) \(\triangle DBA\sim\triangle DAC\)
Hence, options (B), (C) & (D) are correct.
