Chapter 6, Problem 4CR

To determine

To Prove: From the given figure $\overleftrightarrow{OR}\perp \text{ bisect }\overline{PM}$

Explanation of Solution

Given information :

We are given that $\angle OMP\cong \angle OPM,\angle PMR\cong \angle MPR$ ,

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

Proof :

As per the question and we have a kite OPRM. Let $\overleftrightarrow{OR}\text{ bisect }\overline{PM}\text{ at Q }$ as shown in the figure.

In $\Delta OPM$ :

  $\angle OMP\cong \angle OPM$

As two angles are equal, we can say $\Delta OPM$ is an isosceles triangle and as we know sides opposite to the equal angles of an isosceles triangle are equal, we can write:

  $\overline{OM}\cong \overline{OP}\mathrm{..........................................}\text{(i)}$

Similarly, In $\Delta PRM$ :

  $\angle PMR\cong \angle MPR$

As two angles are equal, we can say $\Delta PRM$ is an isosceles triangle and as we know sides opposite to the equal angles of an isosceles triangle are equal, we can write:

  $\overline{RM}\cong \overline{PR}\mathrm{..........................................}\text{(ii)}$

Considering $\Delta OPR\text{ and }\Delta OMR$ we have:

  $\begin{array}{l}\overline{OM}\cong \overline{OP}\mathrm{.................................................}\text{(from i) }\\ \overline{MR}\cong \overline{PR}\mathrm{.................................................}\text{(from ii) }\\ \overline{OR}\cong \overline{OR}\mathrm{...................................................}\text{(common)}\end{array}$

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

  $\Delta OPR\text{ and }\Delta OMR$ are congruent

Or $\Delta OPR\cong \Delta OMR$

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:

  $\angle MOR\cong \angle POR\mathrm{.......................................}\text{(iii)}$

Consider $\Delta MOQ\text{ and }\Delta POQ$ we have:

  $\begin{array}{l}\overline{OM}\cong \overline{OP}\mathrm{.................................................}\text{(from i) }\\ \angle MOR\cong \angle POR\mathrm{.......................................}\text{( from iii)}\\ \overline{OQ}\cong \overline{OQ}\mathrm{...................................................}\text{(common)}\end{array}$

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

  $\Delta MOQ\text{ and }\Delta POQ$ are congruent

Or $\Delta MOQ\cong \Delta POQ$

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:

  $\begin{array}{l}\overline{PQ}\cong \overline{QM}\mathrm{.............................................}\text{(iv)}\\ \angle OQM\cong \angle OQP\end{array}$

Also $\angle OQM\text{ and }\angle OQP$ are adjacent angles and form a linear pair as shown in the figure. So

  $\begin{array}{l}\angle OQM\cong \angle OQP\text{ and }\angle OQM\text{ + }\angle OQP\text{ =180}°\\ \angle OQM\text{ + }\angle OQM\text{ =180}°\\ \Rightarrow \text{2}\angle OQM\text{ = 180}°\\ \Rightarrow \angle OQM\text{ = 90}°\text{ =}\angle OQP\mathrm{...................................}(\text{v)}\end{array}$

Combining the result of (iv) and (v) we can say:

  $\begin{array}{l}\overline{PQ}\cong \overline{QM}\\ \angle OQM\cong \angle OQP\text{ = 90}°\end{array}$

Which proves that $\overleftrightarrow{OR}\perp \text{ bisect }\overline{PM}$

Hence the required result.