question_answer

In the given figure, RQ and TP are perpendicular to PQ, also \[TS\bot PR\] prove that \[\text{ST}\text{.RQ=PS}\text{.PQ}\].

Answer:

In \[\Delta \text{ }RPQ\]

\[\angle 1+\angle 2+\angle 4=180{}^\circ \]

\[\angle 1+\angle 2+90{}^\circ =180{}^\circ \]

\[\angle 1+\angle 2=180{}^\circ -90{}^\circ \]

\[\angle 1=90{}^\circ -\angle 2\]                                             ...(i)

\[\because \]                      \[TP\bot PQ\]

\[\therefore \]               \[\angle TPQ=90{}^\circ \]

\[\Rightarrow \]         \[\angle 2+\angle 3=90{}^\circ \]

\[\angle 3=90{}^\circ -\angle 2\]                                              ?(ii)

From eq. (i) and eq. (ii)

\[\angle 1=\angle 3\]

Now in \[\Delta \text{ }RQP\] and \[\Delta \text{ }PST\]

\[\angle 1=\angle 3\]       [Proved above]

\[\angle 4=\angle 5\]                 [Each \[90{}^\circ \]]

So by AA similarity

\[\Delta \,RQP\sim \Delta \,PST\]

\[\frac{ST}{QP}=\frac{PS}{RQ}\].

\[\Rightarrow \]               \[ST.RQ=PS.PQ\]                      Hence Proved