Chapter 5.1, Problem 13PSB

To determine

To Prove:

  $\begin{array}{l}\text{If }\Delta ABC\text{is isosceles with base }\overline{BC}\text{and if P is a point on }\overline{BC}\text{that is not the midpoint,}\\ \text{then }\overrightarrow{AP}\text{ does not bisect }\angle BAC.\end{array}$ .

Explanation of Solution

Given:

  

  $\Delta ABC\text{is isosceles with base }\overline{BC}\text{and if P is a point on }\overline{BC}\text{that is not the midpoint}$

Concept Used:

In isosceles two sides are of equal measure.

Proof:

Here, we have

  $\Delta ABC\text{is isosceles with base }\overline{BC}\text{and if P is a point on }\overline{BC}\text{that is not the midpoint}$

  $\begin{array}{l}\text{In }\Delta BAP\text{and }\Delta CAP\\ \Rightarrow \overline{BA}\cong \overline{CA}\left(\because \text{Equal sides of isosceles triangle}\right)\\ \Rightarrow \overline{AP}\cong \overline{AP}\left(\because \text{Common side}\right)\\ \Rightarrow \angle ABP\cong \angle CBP\left(\because \text{Angles opposite to side of equal length are equal in measure}\right)\\ \Rightarrow \Delta BAP\cancel{\cong }\Delta CAP\left(\because \text{Triangles are not congruent on the basis of ASS axioms}\right)\\ \Rightarrow \angle BAP\cancel{\cong }\angle CAP\\ \Rightarrow \overrightarrow{AP}\text{ does not bisect }\angle BAC\end{array}$