question_answer

In the figure, there are two points D and E on side AB of \[\Delta \,ABC\] such that \[AD=BE\]. If \[DP\parallel BC\]and \[EQ\parallel AC\], then prove that \[PQ\parallel AB\].

Answer:

In \[\Delta \,ABC\],

\[DP\parallel BC\]                    (Given)

\[\Rightarrow \frac{AD}{DB}=\frac{AP}{PC}\]                 ?(i) [Thales? Theorem]

Also,                   \[EQ\parallel AC\]                                 (Given)

\[\Rightarrow \frac{BE}{EA}=\frac{BQ}{QC}\]                [Thales? Theorem]

\[\Rightarrow \frac{AD}{DB}=\frac{BQ}{QC}\]                ...(ii) \[[\because AD=BE;\,\,\therefore EA=DB]\]

From eq. (i) and (ii)

\[\frac{AP}{PC}=\frac{BQ}{PC}\]

\[\therefore PQ\parallel AB\]                     (Inverse of Thales theorem)          Hence Proved.