Chapter 3.6, Problem 8PSB

To determine

To Prove: In the given figure $\Delta RIB$ is right angled triangle.

Explanation of Solution

Given information : In the figure we are given that $\overline{BI}\cong \overline{RD}\text{ , }\overline{RI}\cong \overline{BD}$ and $\angle 3$ is complement to $\angle 2$

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

Proof: According to the question:

  →$\begin{array}{l}\overline{BI}\cong \overline{RD}\mathrm{.............................................}\text{(i)}\\ \\ \overline{RI}\cong \overline{BD}\mathrm{.........................................................................................}\text{(ii)}\end{array}$

and $\angle 3$ is complement to $\angle 2$ which means $\angle 2\text{ + }\angle 3\text{ = 90}°\mathrm{...................................................................}\text{(iii)}$

Now consider $\Delta RID,\Delta BDI$

  $\begin{array}{l}\overline{BI}\cong \overline{RD}\mathrm{................................................................................}\text{( from i)}\\ \\ \overline{RI}\cong \overline{BD}\mathrm{........................................................................}\text{( from ii)}\\ \\ \overline{DI}\cong \overline{DI}\mathrm{......................................................................}\text{(Common/reflexive)}\end{array}$

Using SSS(Side-Side-Side) rule we can say:

  $\Delta RID\cong \Delta BDI$

As the triangles are congruent thus by CPCT rule their corresponding parts are also congruent. $$

  $\angle \text{1 }\cong \angle \text{3}\mathrm{...............................................................}\text{(iv)}$

Using (iii) and(iv) we get:

  $\begin{array}{l}\angle 2\text{ + }\angle 1\text{ = 90}°\\ \\ \Rightarrow \angle RID\text{ + }\angle \text{DIB = 90}°\\ \\ \Rightarrow \angle \text{RIB = 90}°\end{array}$

As one angle in $\Delta RIB$ is of measure $90°$ ,thus we can say it is a right- angled triangle.

Which proves the required result.