Correct option is (A) BD . DC
Given that \(\angle A=90^\circ\) and \(AD\bot BC\)
Let \(\angle BAD=\angle 1\;\&\;\angle CAD=\angle 2\)
\(\because\) \(\angle A+\angle B+\angle C=180^\circ\)
\(\Rightarrow\) \(\angle B+\angle C=90^\circ\) ________________(1) \((\because\angle A=90^\circ)\)
Now in \(\triangle ABD,\)
\(\angle ADB+\angle B+\angle 1=180^\circ\)
\(\Rightarrow\) \(\angle B+\angle 1=180^\circ-90^\circ=90^\circ\) ________________(2) \((\because\angle ADB=90^\circ\;as\;AD\bot BC)\)
From (1) & (2), we obtain
\(\angle C=\angle 1\) ________________(3)
Now, in triangles \(\triangle ADB\;\&\;\triangle ADC,\)
\(\angle 1=\angle C\) (From (3))
\(\angle ADB=\angle ADC=90^\circ\) \((AD\bot BC)\)
\(\therefore\) \(\triangle ADB\sim\triangle CDA\) (By AA similarity rule)
\(\therefore\) \(\frac{AD}{CD}=\frac{BD}{AD}\)
\(\Rightarrow AD^2=BD.CD=BD.DC\)
