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
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