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.
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