Chapter 6, Problem 2RP

To determine

To Prove: $\Delta ABC$is an isosceles triangle

Explanation of Solution

Given information :

we are given that $\overline{PB}\perp m\text{ , }\overline{PA}\cong \overline{PC}$ ,

Concept used : RHS (Right angle − Hypotenuse- 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:

  $\begin{array}{l}\overline{PB}\perp m\\ \Rightarrow \angle PBA\cong \angle PBC\cong \text{ 90}°\mathrm{...................................}(\text{i)}\\ \text{ also }\overline{PA}\cong \overline{PC}\mathrm{..........................................}\text{(ii)}\end{array}$

Considering $\Delta PBC\text{ and }\Delta PBA$ we have:

  $\begin{array}{l}\overline{PC}\cong \overline{PA}\mathrm{............................................}\text{(from ii)}\\ \overline{PB}\cong \overline{PB}\mathrm{............................................}\text{(common)}\\ \angle PBA\cong \angle PBC\text{ = 90}°\mathrm{............................}\text{(from i)}\end{array}$

So, using RHS (Right angle − Hypotenuse- Side) Congruency rule we can say that:

  $\Delta PBC\text{and }\Delta PBA$ are congruent

Or $\Delta PBC\cong \Delta PBA$

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:

  $\Rightarrow \overline{AB}\cong \overline{BC}$

As two sides of $\Delta ABC$ are equal so we can say that it is an isosceles triangle.

Which proves the required result.