In Δ ABC, line $DE \parallel BC$. If AD = $3x - 2, AE = 5x - 4, BD = 7x - 5$ and CE = $5x - 3$ find the value of x using the Basic Proportionality Theorem (BPT). The solution yields $x = 1$ or $x = \frac{7}{10}$.

Step 1: Apply Basic Proportionality Theorem (BPT)

Since the line segment $DE$ is parallel to the base $BC$ in $\triangle ABC$, we can apply the Basic Proportionality Theorem (also known as Thales Theorem). This theorem states that if a line parallel to one side of a triangle intersects the other two sides, then it divides the two sides proportionally.

Step 2: Substitute the given values

Now, we substitute the given expressions for the lengths of the segments into the proportionality equation. We have $AD = 3x - 2$, $BD = 7x - 5$, $AE = 5x - 4$, and $CE = 5x - 3$.

Step 3: Solve the equation for x

To solve for $x$, we cross-multiply the terms in the equation. Then, we expand both sides and rearrange the terms to form a quadratic equation in the standard form $ax^2 + bx + c = 0$.

Step 4: Factor the quadratic equation

We solve the quadratic equation $10x^2 - 17x + 7 = 0$ by factoring. We look for two numbers that multiply to $10 \times 7 = 70$ and add up to $-17$. These numbers are $-10$ and $-7$. We then split the middle term and factor by grouping.

Step 5: Find the possible values of x

From the factored form $(10x - 7)(x - 1) = 0$, we set each factor equal to zero to find the possible values of $x$. This gives us two potential solutions for $x$.

Step 6: Check for valid lengths

For the lengths of the segments to be valid, they must be positive. We test both values of $x$. When $x = \frac{7}{10}$, $BD$ becomes negative, which is not possible for a length. Therefore, $x = \frac{7}{10}$ is not a valid solution. When $x = 1$, all segment lengths are positive, making it a valid solution.

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