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For triangle ABC,R=5/2 and r=1. Let I be...

For triangle `ABC,R=5/2 and r=1.` Let `I` be the incenter of the triangle and `D,E and F` be the feet of the perpendiculars from `I->BC,CA and AB,` respectively. The value of `(ID*IE*IF)/(IA*IB*IC)` is equal to (a) `5/2` (b) `5/4` (c) `1/10` (d) `1/5`

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To solve the problem, we need to find the value of \(\frac{ID \cdot IE \cdot IF}{IA \cdot IB \cdot IC}\) given that the circumradius \(R = \frac{5}{2}\) and the inradius \(r = 1\). ### Step-by-Step Solution: 1. **Understanding the Problem**: We have a triangle \(ABC\) with incenter \(I\) and circumradius \(R\) and inradius \(r\) given. The points \(D\), \(E\), and \(F\) are the feet of the perpendiculars from \(I\) to the sides \(BC\), \(CA\), and \(AB\) respectively. 2. **Using the Relationship**: ...
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