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D is the mid-point of side BC of $\triangle ABC$. CE and BF intersect at O, a point on AD. AD is produced to G such that $OD = DG$. Prove that
(i) OBGC is a parallelogram.
(ii) $EF \parallel BC$
(iii) $\triangle AEF \sim \triangle ABC$
(i) OBGC is a parallelogram.
(ii) $EF \parallel BC$
(iii) $\triangle AEF \sim \triangle ABC$
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Sol. (i) $\therefore$ Diagonals OG and BC of quadrilateral OBGC bisect each other.
$\therefore$ OBGC is a parallelogram (I Mark)
(ii) $CO \parallel GB \Rightarrow CE \parallel GB$ (II Mark)
In $\triangle AGB, OE \parallel GB \rightarrow \frac{AO}{OG} = \frac{AE}{EB}$ (III Mark)
Similarly in $\triangle AGC, \frac{AO}{OG} = \frac{AF}{FC}$ (IV Mark)
$\Rightarrow \frac{AE}{EB} = \frac{AF}{FC} \Rightarrow EF \parallel BC$ (V Mark)
(iii) In $\triangle AEF$ and $\triangle ABC$
$\angle AEF = \angle ABC$ and $\angle A$ is common.
$\therefore \triangle AEF \sim \triangle ABC$ (VI Mark)
$\therefore$ OBGC is a parallelogram (I Mark)
(ii) $CO \parallel GB \Rightarrow CE \parallel GB$ (II Mark)
In $\triangle AGB, OE \parallel GB \rightarrow \frac{AO}{OG} = \frac{AE}{EB}$ (III Mark)
Similarly in $\triangle AGC, \frac{AO}{OG} = \frac{AF}{FC}$ (IV Mark)
$\Rightarrow \frac{AE}{EB} = \frac{AF}{FC} \Rightarrow EF \parallel BC$ (V Mark)
(iii) In $\triangle AEF$ and $\triangle ABC$
$\angle AEF = \angle ABC$ and $\angle A$ is common.
$\therefore \triangle AEF \sim \triangle ABC$ (VI Mark)