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Answer variants:
\(BD\)
\(\triangle FEC \cong \triangle GBD\)
[Since corresponding angles are equal]
\(\angle A\)
\(\angle 1\)
\(AE\)
\(DE \parallel BC\)
Using the data and the figure given below, prove that \(\triangle ADE \sim \triangle ABC\).
(i) \(\triangle FEC \cong \triangle GBD\)
(ii) \(\angle 1 = \angle 2\)
Solution:
We know that .
Thus, \(FC\) \(=\) \(\longrightarrow (1)\)
We also know that \(=\) \(\angle 2\).
By , "The sides opposite to equal angles of a triangle are equal."
Thus, \(=\) \(AD\) \(\longrightarrow (2)\)
On equating \((1)\) and \((2)\), we have:
\(\frac{AE}{EC}\) \(=\)
By , "If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side."
So, .
Also, \(\angle 1 = \angle 3\), and \(\angle 2 = \angle 4\).
Also, is common to both \(\triangle ADE\) and \(\triangle ABC\).
By similarity criterion, "If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar."
Thus, \(\triangle ADE \sim \triangle ADC\).
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