Chapter 8, Problem 15CUR

To determine

To Prove: The ΔABC is isosceles in which AX is median and an altitude.

Explanation of Solution

Given information:

Consider the ΔABC.

  

The figure is shown below;

  

Since AX is median and an altitude of ΔABC.

This implies that,

  $\begin{array}{l}\angle AXC=\angle AXB\left(={90}^{\circ }\right)\hfill \\ BX=XC\hfill \end{array}$

In $\text{Δ}AXB$ and $\text{Δ}AXC$

  $\begin{array}{cc}\angle AXC& =\angle AXB\left(={90}^{\circ }\right)\\ BX& =XC\\ AX& =AX(\text{common})\end{array}$

Hence $\text{Δ}AXB\cong \text{Δ}AXC\text{So}AB=AC,\text{and}\text{Δ}AXB$ is isosceles.