Chapter 3.6, Problem 9PSB

To determine

To Conclude: In the given figure $\Delta EHJ$ is an isosceles

Explanation of Solution

Given information : We are given $\angle EFJ\cong \angle HGJ\text{ , }\overline{JF}\cong \overline{JG}$ , $F$ and $G$ trisect $\overline{EH}$ .

Formula used : Congruency criteria of SAS and CPCT (corresponding parts of congruent triangles) rule.

Proof: As per the given question we have $\overline{JF}\cong \overline{JG}\mathrm{.............................................}\text{(i)}$

  $\angle EFJ\cong \angle HGJ\mathrm{............................................................................}\text{(ii)}$

  $F$ and $G$ trisect $\overline{EH}$ that means it divides $\overline{EH}$ in three equal parts .So we can write it as:

  $\overline{EF}\cong \overline{FG}\cong \overline{GH}\mathrm{.........................................................................}\text{(iii)}$

Consider $\Delta EFJ\text{ ,}\Delta HGJ$

  $\begin{array}{l}\overline{JF}\cong \overline{JG}\mathrm{.................................................................}\text{(from i)}\\ \\ \angle EFJ\cong \angle HGJ\mathrm{..........................................................}\text{(from ii)}\\ \\ \overline{EF}\cong \overline{GH}\mathrm{...............................................................}\text{(from iii)}\end{array}$

Thus, by SAS(Side-Angle-Side) criteria we can say:

  $\Delta EFJ\cong \Delta HGJ$

As the triangles are congruent thus by CPCT rule their corresponding parts also will be congruent $\overline{JE}\cong \overline{HJ}$

Now for triangle EHJ two of its sides are equal so the triangle is an isosceles triangle.

Which proves the required result.