issue a notice in the form of letter to the members of your housing society inviting them to attend a lecture on ' save water ' by an eminent environmentalist based on the following points .
⚪ severe water shortage in pune for the past year.
⚪ serious measures necessary.
⚪lecture by eminent environmentalist ( name ) venue : society .
⚪ Date :- 1st may 2023 . time : 5 pm
Answers & Comments
Answer:
Solution−
Given that,
\begin{gathered}\sf \: In\:\triangle\:ABC \\ \\ \end{gathered}
In△ABC
E is the midpoint of median AD and BE is produced to meet AC at F.
As given that, AD is median of triangle ABC.
\begin{gathered}\sf\implies \sf \: D \: is \: midpoint \: of \: BC \\ \\ \end{gathered}
⟹DismidpointofBC
Construction: Through D, draw a line DG parallel to BF intersecting AC at G.
Now,
\begin{gathered}\sf \: In\:\triangle\:ADG \\ \\ \end{gathered}
In△ADG
\begin{gathered} \sf \: E \: is \: midpoint \: of \: AD \\ \\ \end{gathered}
EismidpointofAD
and
\begin{gathered}\sf \: EF \: \parallel \: DG \\ \\ \end{gathered}
EF∥DG
By Converse of Midpoint Theorem,
\begin{gathered}\sf\implies \sf \: F \: is \: the \: midpoint \: of \: AG \\ \\ \end{gathered}
⟹FisthemidpointofAG
\begin{gathered}\sf\implies \sf \: AF = FG - - - (1) \\ \\ \end{gathered}
⟹AF=FG−−−(1)
Now,
\begin{gathered}\sf \: In\:\triangle\:BFC \\ \\ \end{gathered}
In△BFC
\begin{gathered} \sf \: D \: is \: midpoint \: of \: BC \\ \\ \end{gathered}
DismidpointofBC
and
\begin{gathered}\sf \: BF \: \parallel \: DG \\ \\ \end{gathered}
BF∥DG
By Converse of Midpoint Theorem,
\begin{gathered}\sf\implies \sf \: G \: is \: the \: midpoint \: of \: FC \\ \\ \end{gathered}
⟹GisthemidpointofFC
\begin{gathered}\sf\implies \sf \: FG = GC - - - (2) \\ \\ \end{gathered}
⟹FG=GC−−−(2)
From equation (1) and (2), we concluded that
\begin{gathered}\sf\implies \sf \:AF = FG = GC \\ \\ \end{gathered}
⟹AF=FG=GC
\begin{gathered}\sf\implies \sf \:AF = \dfrac{1}{3}AC\\ \\ \end{gathered}
⟹AF=
3
1
AC
\rule{190pt}{2pt}Solution−
Given that,
\begin{gathered}\sf \: In\:\triangle\:ABC \\ \\ \end{gathered}
In△ABC
E is the midpoint of median AD and BE is produced to meet AC at F.
As given that, AD is median of triangle ABC.
\begin{gathered}\sf\implies \sf \: D \: is \: midpoint \: of \: BC \\ \\ \end{gathered}
⟹DismidpointofBC
Construction: Through D, draw a line DG parallel to BF intersecting AC at G.
Now,
\begin{gathered}\sf \: In\:\triangle\:ADG \\ \\ \end{gathered}
In△ADG
\begin{gathered} \sf \: E \: is \: midpoint \: of \: AD \\ \\ \end{gathered}
EismidpointofAD
and
\begin{gathered}\sf \: EF \: \parallel \: DG \\ \\ \end{gathered}
EF∥DG
By Converse of Midpoint Theorem,
\begin{gathered}\sf\implies \sf \: F \: is \: the \: midpoint \: of \: AG \\ \\ \end{gathered}
⟹FisthemidpointofAG
\begin{gathered}\sf\implies \sf \: AF = FG - - - (1) \\ \\ \end{gathered}
⟹AF=FG−−−(1)
Now,
\begin{gathered}\sf \: In\:\triangle\:BFC \\ \\ \end{gathered}
In△BFC
\begin{gathered} \sf \: D \: is \: midpoint \: of \: BC \\ \\ \end{gathered}
DismidpointofBC
and
\begin{gathered}\sf \: BF \: \parallel \: DG \\ \\ \end{gathered}
BF∥DG
By Converse of Midpoint Theorem,
\begin{gathered}\sf\implies \sf \: G \: is \: the \: midpoint \: of \: FC \\ \\ \end{gathered}
⟹GisthemidpointofFC
\begin{gathered}\sf\implies \sf \: FG = GC - - - (2) \\ \\ \end{gathered}
⟹FG=GC−−−(2)
From equation (1) and (2), we concluded that
\begin{gathered}\sf\implies \sf \:AF = FG = GC \\ \\ \end{gathered}
⟹AF=FG=GC
\begin{gathered}\sf\implies \sf \:AF = \dfrac{1}{3}AC\\ \\ \end{gathered}
⟹AF=
3
1
AC
\rule{190pt}{2pt}