**Prove that if PR is not parallel to BC, then \(\frac{|AP|}{|AB|} \neq \frac{|AR|}{|AC|}\) (Figure 1.13).**If P is any point on AB and Q is any point on AC, then PQ is parallel to BC *if and only if* \(\frac{|AP|}{|AB|} = \frac{|AQ|}{|AC|}\). *(Figure 1.13)*(Note: The text contains a typographical error, "paraellel" should be "parallel.") The image is a diagram illustrating the "Converse of the Thales Theorem," labeled as Figure 1.13.### Diagram Details:- **Triangle **: The diagram shows triangle \( ABC \).- **Points and Lines**: - A line \( \ell \) extends across the diagram with points labeled \( P \) and \( Q \) on it. - Point \( A \) is the vertex of the triangle opposite the base \( BC \). - Two segments, \( AP \) and \( AQ \), originate from point \( A \) extending downwards intersecting line \( \ell \) at \( P \) and \( Q \). - Another segment \( PR \) extends from \( P \) and intersects the triangle at point \( R \) on side \( AC \).This diagram is typically used to demonstrate the converse of the Thales Theorem, which pertains to the properties of a line parallel to one side of the triangle and its intersections with the other two sides.

Prove that:if PR is not parallel to BC then .Given statement isif P is any point on AB and Q is any…

Prove that if PR is not parallel to BC then # figure 1.13 AP |AR| |AB| |AC| . |AP| If P is any point on AB and Q is any point on AC then PQ is paraellel to BC if and only if 1.13* |AB| |AQ| = *figure |AC|

Prove that if PR is not parallel to BC then # figure 1.13 AP |AR| |AB| |AC| . |AP| If P is any point on AB and Q is any point on AC then PQ is paraellel to BC if and only if 1.13* |AB| |AQ| = *figure |AC|

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Question

**Prove that if PR is not parallel to BC, then \(\frac{|AP|}{|AB|} \neq \frac{|AR|}{|AC|}\) (Figure 1.13).**

If P is any point on AB and Q is any point on AC, then PQ is parallel to BC *if and only if* \(\frac{|AP|}{|AB|} = \frac{|AQ|}{|AC|}\). *(Figure 1.13)*

(Note: The text contains a typographical error, "paraellel" should be "parallel.")

The image is a diagram illustrating the "Converse of the Thales Theorem," labeled as Figure 1.13.

### Diagram Details:

- **Triangle **: The diagram shows triangle \( ABC \).
- **Points and Lines**:
- A line \( \ell \) extends across the diagram with points labeled \( P \) and \( Q \) on it.
- Point \( A \) is the vertex of the triangle opposite the base \( BC \).
- Two segments, \( AP \) and \( AQ \), originate from point \( A \) extending downwards intersecting line \( \ell \) at \( P \) and \( Q \).
- Another segment \( PR \) extends from \( P \) and intersects the triangle at point \( R \) on side \( AC \).

This diagram is typically used to demonstrate the converse of the Thales Theorem, which pertains to the properties of a line parallel to one side of the triangle and its intersections with the other two sides.

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