Question

$\Delta ABC$ and $\Delta DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the opposite side of $BC$ such that $AB=AC$ and $DB=DC$.

Show that $AD$ is the perpendicular bisector of $BC$.

Answer 1

Step $1:$ Drawing the diagram:

$\Delta ABC$ and $\Delta DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the opposite side of $BC$, such that

$AB=AC$ and $DB=DC$

Join the diagonals $AD$ and $BC$ which intersect at $O$ and marked angles as $\angle 1,\angle 2,\angle 3,\angle 4$.

Step $2:$ Proving $\angle 1=\angle 2$:

In $\Delta ABD$ and $\Delta ACD$

$$\begin{gathered} AB=AC\left(\mathrm{Given}\right) \\ BD=CD\left(\mathrm{Given}\right) \\ AD=AD(\mathrm{Common}\mathrm{side}) \\ \therefore ∆ABD\cong ∆ACD(\mathrm{By}\mathrm{S}.\mathrm{S}.\mathrm{S}\mathrm{congruency}) \end{gathered}$$

$$\begin{gathered} \therefore \angle BAD=\angle CAD(\mathrm{By}\mathrm{C}.\mathrm{P}.\mathrm{C}.\mathrm{T}) \\ \Rightarrow \angle 1=\angle 2...................(\mathrm{equation}1) \end{gathered}$$

Step $3:$ Proving $AD$ is the perpendicular bisector of $BC$:

In $\Delta OAB$ and $\Delta OAC$

$\begin{array}{rcl}AB& =& AC\left(\mathrm{Given}\right)\\ \angle BAO& =& \angle CAO\left(\mathrm{From}\mathrm{equation}1\right)\\ AO& =& AO(\mathrm{Common}\mathrm{side})\end{array}$

$\therefore \Delta OAB\cong \Delta OAC\left(\mathrm{By}\mathrm{SAS}\mathrm{congruency}\right)$

$\begin{array}{rcl}\therefore \angle AOB& =& \angle AOC\left(\mathrm{By}\mathrm{C}.\mathrm{P}.\mathrm{C}.\mathrm{T}\right)\\ OB& =& OC\left(\mathrm{By}\mathrm{C}.\mathrm{P}.\mathrm{C}.\mathrm{T}\right)\end{array}$

$\begin{array}{rcl}\angle AOB+\angle AOC& =& 180°\\ 2\angle AOB& =& 180°\\ \angle AOB& =& 90°=\angle AOC\end{array}$

Therefore, $AD$ is the perpendicular bisector of $BC$.

Hence proved.