Chapter 6, Problem 3RP

To determine

To Prove: From the given figure $\overline{DB}\cong \overline{DC}$

Explanation of Solution

Given information : we are given that $\overline{AB}\cong \overline{AC}\text{ , }\angle DAB\cong \angle DAC$ ,

Concept used : SAS (Side-Angle- Side) Congruency and CPCT (corresponding parts of congruent triangles are equal) rule.

Proof :

As per the question and we have a plane and triangle placed over it where:

  $\overline{AB}\cong \overline{AC}\text{ , }\angle DAB\cong \angle DAC$

Considering $\Delta DAB\text{ and }\Delta DAC$ we have:

  $\begin{array}{l}\overline{AB}\cong \overline{AC}\mathrm{.................................................}\text{(given) }\\ \angle DAB\cong \angle DAC\mathrm{.........................................}\text{(given) }\\ \overline{DA}\cong \overline{DA}\mathrm{...................................................}\text{(common)}\end{array}$

So, using SAS (Side-Angle- Side) Congruency rule we can say that:

  $\Delta DAB\text{ and }\Delta DAC$ are congruent

Or $\Delta DAB\cong \Delta DAC$

As we know that when two triangles are congruent then by CPCT rule their corresponding parts are also equal so we can say for these two triangles we have:

  $\Rightarrow \overline{DB}\cong \overline{DC}$

Which proves the required result.