Chapter 12, Problem 22CR

To determine

To prove: D is the midpoint of $\overleftrightarrow{AC}$ in $△ABC$

Explanation of Solution

Given information: Given, $\overleftrightarrow{AC}\perp m$ and $\overleftrightarrow{BC}\equiv \overleftrightarrow{BA}$

Formula used:

Side-Angle-Side (SAS) theorem of congruency of triangles.

Calculation:

In $△BDA$ and $△BDC$

  $\overleftrightarrow{BA}\equiv \overleftrightarrow{BC}$ (Given in the problem)

  $\angle BDA=\angle BDC$ (As it is given in the problem that $\overleftrightarrow{AC}\perp m$ )

  $\overleftrightarrow{BD}=\overleftrightarrow{BD}$ ( $\overleftrightarrow{BD}$ is a side which is common to both the triangles)

So, by Side-Angle-Side (SAS) theorem of congruency of triangles

  $△BDA\cong △BDC$

Then, $\overleftrightarrow{DA}=\overleftrightarrow{DC}$ (Corresponding Parts of Congruent Triangles (CPCT))

Hence, D is the midpoint of $\overleftrightarrow{AC}$ in $△ABC$ .