Question

31. In Fig 7.41, ABCD is a parallelogram in which Eis
the point on BC. On producing DE and AB meets at
F. If AF=2AB, prove that E is the mid-point of BC.
​

Answer 1

$\begin{gathered}\begin{gathered}\bf Given - \begin{cases} &\sf{ ABCD \: is \: a \: parallelogram} \\ &\sf{DE \: and \: AB \: meets \: at \: </p><p>F.}\\ &\sf{E \: is \: </p><p>the \: point \: on \: BC}\\ &\sf{AF \: = \: 2AB} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf To \: prove :- \begin{cases} &\sf{E \: is \: the \: mid-point \: of \: BC.} \end{cases}\end{gathered}\end{gathered}$

Proof :-

$\tt \:  Since \: ABCD \: is \: a \: parallelogram. \: \: \: \: \: \\ \tt \:  </u></p><p><u>[tex]\tt \:  Since \: ABCD \: is \: a \: parallelogram. \: \: \: \: \: \\ \tt \:  We \: know, \: opposite \: sides \: are \: equal \\ \tt \:  ⟼ \: </u></p><p><u>[tex]\tt \:  Since \: ABCD \: is \: a \: parallelogram. \: \: \: \: \: \\ \tt \:  We \: know, \: opposite \: sides \: are \: equal \\ \tt \:  ⟼ \: Therefore, \: AB \: = CD - - - -(1)$

$\tt \:  ⟼Now, AF = 2AB \:\: \: \: \: \: \\ </u></p><p><u>[tex]\tt \:  ⟼Now, AF = 2AB \:\: \: \: \: \: \\ \tt \:  :⟹ \: AB + BF = 2AB \: \: \: \: \: \\ \tt \:  </u></p><p><u>[tex]\tt \:  ⟼Now, AF = 2AB \:\: \: \: \: \: \\ \tt \:  :⟹ \: AB + BF = 2AB \: \: \: \: \: \\ \tt \:  :⟹ \ BF = 2AB - AB \: \: \: \: \: \: \\ \tt \:  </u></p><p><u>[tex]\tt \:  ⟼Now, AF = 2AB \:\: \: \: \: \: \\ \tt \:  :⟹ \: AB + BF = 2AB \: \: \: \: \: \\ \tt \:  :⟹ \ BF = 2AB - AB \: \: \: \: \: \: \\ \tt \:  :⟹ BF = AB - --(2)$

$\tt \:  ⟼ From \: (1) \: and \: (2), \: we \: conclude \: that$

$\tt \:  ⟼ \: BF = CD ---(3)$

$\tt \:  ⟼Now, \: In \: \triangle \: CDE \: and \: \triangle \: EBF$

$\tt \:  ⟼ ∠CED = ∠BEF --(Vertically \: opposite \: angles) \\\tt \:  </u></p><p><u>[tex]\tt \:  ⟼ ∠CED = ∠BEF --(Vertically \: opposite \: angles) \\\tt \:  ⟼ ∠ECD = ∠EBF --(Alternate \: interior \: angles ) \\ \tt \: </u></p><p><u>[tex]\tt \:  ⟼ ∠CED = ∠BEF --(Vertically \: opposite \: angles) \\\tt \:  ⟼ ∠ECD = ∠EBF --(Alternate \: interior \: angles ) \\ \tt \: ⟼ BF = AB --(Proved above (3)) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \tt \:  ⟼∆CDE ≅ ∆BFE ---(SAS \: Congruency \: Rule) \: \: \: \: \: \:$

$\tt \:  :⟹ CE = BE ----- (CPCT) \\ \tt \:  </u></p><p><u>[tex]\tt \:  :⟹ CE = BE ----- (CPCT) \\ \tt \:  :⟹ E \: is \: the \: midpoint \: of \: BC. \: \: \: \:$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$