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Chapter 5: Problem 17

Point \(B\) lies inside the segment \([A C] .\) Equilateral triangles \(A B E\) and \(B C F\) are constructed on the same side of line \(A C .\) If \(M\) and \(N\) are the midpoints of segments \(A F\) and \(C E\), prove that triangle \(B M N\) is equilateral.

Short Answer

Expert verified
Based on the given solution, create a short answer question: Question: Prove that triangle BMN is equilateral given that triangles ABE and BCF are equilateral and M and N are the midpoints of segments AF and CE, respectively. Answer: To prove that triangle BMN is equilateral, we need to show that all its sides have equal length. We first find the lengths of segments AM and AN, using the midpoints and side lengths of triangles ABE and BCF. Next, we prove that triangle AMC is equilateral by comparing the sides AE + AN and CF + AM to AC, and find that these sides are equal. Then, we find the lengths of sides BN, BM, and MN, and show that they are all equal, proving that triangle BMN is equilateral.

Step by step solution

01

Determine side lengths of the given triangles

Since \(ABE\) and \(BCF\) are equilateral triangles, we know that \(AB = BE = AE\) and \(BC = BF = CF\). Let the side length of triangle \(ABE\) be \(x\) and the side length of triangle \(BCF\) be \(y\). Therefore, \(AB = BE = AE = x\), and \(BC = BF = CF = y\).
02

Determine the length of the segment AC

We know that point \(B\) lies inside the segment \(AC\). Therefore, we can find the length of segment \(AC\) by adding the lengths of segments \(AB\) and \(BC\). So, \(AC = AB + BC = x + y\).
03

Determine the lengths of segments \(AM\) and \(AN\)

Since \(M\) and \(N\) are the midpoints of segments \(AF\) and \(CE\), we can say that \(AM = \frac{AF}{2}\) and \(AN = \frac{CE}{2}\). We know that \(AF = AC + CF = (x + y) + y = x + 2y\) and \(CE = CA + AE = (x + y) + x = 2x + y\). Therefore, \(AM = \frac{x + 2y}{2}\) and \(AN = \frac{2x + y}{2}\).
04

Prove that triangle \(AMC\) is equilateral

We know that \(AE = x\) and \(CF = y\). From Step 3, we have \(AN = \frac{2x + y}{2}\) and \(AM = \frac{x + 2y}{2}\). By adding \(AE\) and \(AN\), we get \(AE + AN = x + \frac{2x + y}{2} = \frac{3x + y}{2}\). Similarly, by adding \(CF\) and \(AM\), we get \(CF + AM = y + \frac{x + 2y}{2} = \frac{x + 3y}{2}\). Since \(AC = x + y\) and \(AE + AN = CF + AM\), we can say that triangle \(AMC\) is equilateral and its side length is \(\frac{3x + 3y}{2}\).
05

Prove that triangle \(BMN\) is equilateral by showing its sides have equal length

Now, let's find the lengths of the sides of triangle \(BMN\): \(BN\), \(BM\) and \(MN\). We have \(BN = BE + EN = x + \frac{x + y}{2}\). We also have \(BM = BF + FM = y + \frac{x + y}{2}\). To find \(MN\), we subtract \(AM\) from \(AN\) such that \(MN = AN - AM = \frac{y}{2}\). The length of sides \(BN\) and \(BM\) is the same: \(BN = BM = x + \frac{x + y}{2}\). Since triangle \(AMC\) is equilateral, we have \(MC = \frac{3x + 3y}{2}\). Therefore, \(MN = MC - (AM + AN) = \frac{3x + 3y}{2} - (\frac{x + 2y}{2} + \frac{2x + y}{2}) = x + \frac{x + y}{2}\). Now, we have proved that all the sides of triangle \(BMN\) have equal length: \(BN = BM = MN = x + \frac{x + y}{2}\). Therefore, triangle \(BMN\) is equilateral.

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