Proving AB = BC = DE using the properties of collinearity and the given information about AC and BD-CE

If A, B, C, D, and E are collinear with B between A and C, C between B and D, and D between Cand E, and AC = BD – CE, then AB = BC = DE

To prove that AB = BC = DE, we will use the properties of collinearity and the fact that AC = BD – CE

To prove that AB = BC = DE, we will use the properties of collinearity and the fact that AC = BD – CE.

First, let’s consider the points A, B, C, D, and E on a line. We are given that B is between A and C, C is between B and D, and D is between C and E. This information tells us the order in which the points are arranged on the line.

Now, let’s use the property of collinearity. A, B, and C are collinear, so the distance AB + BC should be equal to the distance AC.

AC = AB + BC ——- (Equation 1)

We are also given that AC = BD – CE. Substituting this value in equation 1, we get:

BD – CE = AB + BC

Rearranging the terms, we have:

AB + BC = BD – CE ——- (Equation 2)

Now, let’s consider the points B, C, D, and E. We are given that C is between B and D, and D is between C and E. Using the property of collinearity, the distance BC + CD should be equal to the distance BD.

BD = BC + CD ——- (Equation 3)

We also know that AC = BD – CE, which means that BD = AC + CE. Substituting this value in equation 3, we get:

AC + CE = BC + CD

Rearranging the terms, we have:

AC + BC = CD + CE ——- (Equation 4)

Now, let’s combine equations 2 and 4:

AB + BC = BD – CE
AC + BC = CD + CE

Adding these two equations together, we get:

AB + 2BC + AC = BD – CE + CD + CE

Simplifying the equation, we have:

AB + 2BC + AC = BD + CD

Since D is between C and E, CD = DE. Rearranging the equation, we have:

AB + 2BC + AC = BD + DE

Since AB + BC = AC, we can substitute this value in the equation:

AB + BC + BC = BD + DE

Simplifying the equation, we get:

AB + 2BC = BD + DE

Since BD = BC + CD and CD = DE, we can substitute these values:

AB + 2BC = BC + BC + DE

Simplifying the equation, we have:

AB + 2BC = 2BC + DE

Subtracting 2BC from both sides, we get:

AB = DE

Therefore, we have proved that AB = DE.

To conclude, using the properties of collinearity and the given information AC = BD – CE, we have shown that AB = DE.

Additionally, we can recall that we proved earlier that AB + BC = AC. Since AB = DE, substituting this value in the equation gives us DE + BC = AC. Since AB = BC, we can also conclude that BC = DE.

Hence, AB = BC = DE.

More Answers:

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Understanding the One Third Way Point Formula in Coordinate Geometry: Step-by-Step Explanation and Example
Proving the Statement: Equidistant Points from a Third Point are Collinear

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