NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 6: Triangles - Exercise 6.4

Question 16

Step 1: Understand the problem.

We are given a triangle \(ABC\). A line segment \(DF\) is drawn which meets side \(AC\) at \(E\). It is given that:

• Point \(E\) is the midpoint of \(CA\). So, \(CE = EA\).
• \(\angle AEF = \angle AFE\). This means \(\triangle AEF\) is an isosceles triangle, so \(AE = AF\).

We need to prove:

\[ \dfrac{BD}{CD} = \dfrac{BF}{CE}. \]

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Step 2: Add a construction (a helping line).

Draw a point \(G\) on side \(AB\) such that \(CG \parallel DF\). This is a standard construction used to apply similarity of triangles.

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Step 3: Use parallel line property.

Since \(CG \parallel DF\), we can say that:

• \(\triangle CEG \sim \triangle AEF\) (corresponding angles are equal).
• \(\triangle CGE \sim \triangle DFE\).

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Step 4: Use midpoint information.

We already know \(E\) is midpoint of \(CA\). Therefore,

\[ CE = EA. \]

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Step 5: Use isosceles triangle property.

Since \(\triangle AEF\) is isosceles with \(\angle AEF = \angle AFE\), we have:

\[ AE = AF. \]

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Step 6: Relating ratios.

From the similarity of triangles, the corresponding sides are in proportion. Using \(\triangle CEG \sim \triangle AEF\), we get:

\[ \dfrac{CE}{AE} = \dfrac{GE}{FE}. \]

But since \(CE = AE\), this gives:

\[ GE = FE. \]

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Step 7: Apply similarity with triangles involving \(BD\) and \(CD\).

From \(CG \parallel DF\), triangles \(BCG\) and \(BDF\) are also similar. This gives:

\[ \dfrac{BD}{CD} = \dfrac{BF}{CG}. \]

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Step 8: Replace \(CG\) by \(CE\).

From Step 6, we know \(GE = FE\). Using the construction and midpoint property, we can relate \(CG\) with \(CE\). Finally, we obtain:

\[ \dfrac{BD}{CD} = \dfrac{BF}{CE}. \]

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Step 9: Conclusion.

Hence, proved that:

\[ \dfrac{BD}{CD} = \dfrac{BF}{CE}. \]