(a) BD.CD = BZC² (b) AB.AC = BC² (c) BD.CD=AD² (d) AB.AC =AD²
Solution:
c) BD.CD=AD²
Clarification:
From ∆ADB and ∆ADC,
As indicated by the inquiry, we have,
\[\begin{array}{*{35}{l}}
~ \\
\angle D\text{ }=\angle D\text{ }=\text{ }90{}^\circ (\because AD\bot BC) \\
\end{array}\]
\[\angle DBA\text{ }=\angle DAC\text{ }[each\text{ }point\text{ }=\text{ }90{}^\circ -\angle C]\]
Utilizing AAA closeness standards,
\[ADB\sim ADC\]
\[BD/AD\text{ }=\text{ }AD/CD\]
\[BD.CD\text{ }=\text{ }AD2\]