**Problem Statement:**32) In triangle ABC, D is a point on AB and E is a point on AC such that $ \overline{DE} \parallel \overline{BC} $. If $AD = 2$, $DB = x - 1$, $AE = x$, and $EC = x + 2$, find $AE$.**Solution Explanation:**Since $ \overline{DE} \parallel \overline{BC} $, by the Basic Proportionality Theorem (or Thales's theorem), we have:\[\frac{AD}{DB} = \frac{AE}{EC}\]Given:- $ AD = 2 $- $ DB = x - 1 $- $ AE = x $- $ EC = x + 2 $Substitute the given values into the proportion:\[\frac{2}{x - 1} = \frac{x}{x + 2}\]Cross-multiplying to solve for $x$:\[2(x + 2) = x(x - 1)\]\[2x + 4 = x^2 - x\]Rearrange into a standard quadratic equation form:\[x^2 - 3x - 4 = 0\]Factorizing the quadratic equation:\[(x - 4)(x + 1) = 0\]So, $x$ can be either $4$ or $-1$.Since lengths cannot be negative, $x = 4$.Therefore, the length of $AE$ is:\[AE = x = 4\]

First I have drawn the diagram and then by using definition of similar triangles find the required…

Answered: 32) In triangle ABC, D is a point on AB…

32) In triangle ABC, D is a point on AB and E is a point on AC such that DE || BC. If AD = 2, DB = x - 1, AE = x, and EC = x + 2, find AE.

Algebra and Trigonometry (MindTap Course List)

4th Edition

ISBN:9781305071742

Author:James Stewart, Lothar Redlin, Saleem Watson

Publisher:James Stewart, Lothar Redlin, Saleem Watson

Chapter5: Trigonometric Functions: Right Triangle Approach

Section5.1: Angle Measure

Problem 4E: Object A is travelling along a circle of radius 2, and Object B is travelling along a circle of...

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Question

**Problem Statement:**

32) In triangle ABC, D is a point on AB and E is a point on AC such that $ \overline{DE} \parallel \overline{BC} $. If $AD = 2$, $DB = x - 1$, $AE = x$, and $EC = x + 2$, find $AE$.

**Solution Explanation:**

Since $ \overline{DE} \parallel \overline{BC} $, by the Basic Proportionality Theorem (or Thales's theorem), we have:

\[\frac{AD}{DB} = \frac{AE}{EC}\]

Given:
- $ AD = 2 $
- $ DB = x - 1 $
- $ AE = x $
- $ EC = x + 2 $

Substitute the given values into the proportion:

\[\frac{2}{x - 1} = \frac{x}{x + 2}\]

Cross-multiplying to solve for $x$:

\[2(x + 2) = x(x - 1)\]
\[2x + 4 = x^2 - x\]

Rearrange into a standard quadratic equation form:

\[x^2 - 3x - 4 = 0\]

Factorizing the quadratic equation:

\[(x - 4)(x + 1) = 0\]

So, $x$ can be either $4$ or $-1$.

Since lengths cannot be negative, $x = 4$.

Therefore, the length of $AE$ is:

\[AE = x = 4\]

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