Question

$ \Delta ABC$ and $ \Delta DBC$ are two isosceles triangles on the same base $ BC$ and vertices $ A$ and $ D$ are on the same side of $ BC$. If $ AD$ is extended to intersect $ BC$ at $ P$, show that

Answer 1

Step $1:$ Explaining the diagram:

$\Delta ABC$ and $\Delta DBC$ are two isosceles triangles such that

$AB=AC,DB=DC$

Step $2:$ Proving $\Delta ABD\cong \Delta ACD$:

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

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

$$\begin{gathered} \therefore \angle BAD=\angle CAD(\mathrm{By}\mathrm{C}.\mathrm{P}.\mathrm{C}.\mathrm{T}) \\ \Rightarrow \angle BAP=\angle CAP............\left(1\right) \end{gathered}$$

Step $3:$ Proving $\Delta ABP\cong \Delta ACP$:

In $\Delta ABP$ and $\Delta ACP$

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

$\therefore \Delta ABP\cong \Delta ACP\left(\mathrm{By}\mathrm{SAS}\mathrm{congruency}\right)$

Step $4:$ Proving $AP$ bisects $\angle A$ as well as $\angle D$:

From equation $1$, $\angle BAP=\angle CAP$,

Therefore, $AP$ bisects $\angle A$.

As, we know that, $\Delta ABP\cong \Delta ACP$,

$\therefore BP=CP\left(\mathrm{By}\mathrm{C}.\mathrm{P}.\mathrm{C}.\mathrm{T}\right)..........\left(2\right)$

Now, In $\Delta BDP$ and $\Delta CDP$

$$\begin{gathered} BD=CD\left(\mathrm{Given}\right) \\ BP=CP\left(\mathrm{From}\mathrm{equation}2\right) \\ DP=DP(\mathrm{Common}\mathrm{side}) \\ \therefore ∆BDP\cong ∆CDP(\mathrm{By}\mathrm{S}.\mathrm{S}.\mathrm{S}\mathrm{congruency}) \end{gathered}$$

$\therefore \angle BDP=\angle CDP\left(\mathrm{By}\mathrm{C}.\mathrm{P}.\mathrm{C}.\mathrm{T}\right)$

Therefore, $AP$ bisects $D$.

Step $5:$ Proving $AP$ is the perpendicular bisector of $BC$:

As, we know that, $\Delta ABP\cong \Delta ACP$,

$\therefore \angle APB=\angle APC\left(\mathrm{By}\mathrm{C}.\mathrm{P}.\mathrm{C}.\mathrm{T}\right)$

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

And $BP=CP$

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

Hence, it is proved that

1. $\Delta ABD\cong \Delta ACD$
2. $\Delta ABP\cong \Delta ACP$
3. $AP$ bisects $\angle A$ as well as $\angle D$
4. $AP$ is the perpendicular bisector of $BC$