Chapter 3.6, Problem 15PSC

To determine

To Prove: For the given conditions $\overline{AR}\cong \overline{PB}$

Explanation of Solution

Given information :

We have an obtuse $\Delta PQR$ with the longest side $\overline{PR}$ . $APQ\text{ and }BQR$ are two equilateral triangles lying outside the given triangle. Also, the measure of each angle of equilateral triangles is given as $60°$

Formula used : Measure of all the sides of an equilateral triangle is equal.

Proof :

As per the question we have: $\Delta PQR$ with the longest side $\overline{PR}$ which means $\angle PQR$ is the largest angle of the triangle .

Also, as we want to find $\overline{AR}\text{ and }\overline{BP}$ it means $A,R\text{ and }B,P$ are collinear points

Or more specifically we can say $A,Q,R\text{ and }B,Q,P$ are collinear sets.

Here we can have two cases. First case when $\Delta PQR$ is a scalene triangle and second case when it is isosceles triangle.

First Case: $\Delta PQR$ is scalene triangle:

As equilateral $\Delta APQ$ is made on $\overline{PQ}\text{ of }\Delta PQR$ :

  $\overline{AP}\cong \overline{PQ}\cong \overline{QA}\text{ = }y(\text{let say) }\mathrm{.....................................}(\text{i)}$

Similarly for equilateral $\Delta BQR$ is made on $\overline{QR}\text{ of }\Delta PQR$ : $$

  $\overline{BQ}\cong \overline{QR}\cong \overline{RB}\text{ = }x\text{ (let say)}\mathrm{.................................}\text{(ii)}$

Taking values from (i) and (ii) to find $\overline{AR}\text{ and }\overline{BP}$ :

  $\begin{array}{l}\overline{AR}=\overline{AQ}\text{ + }\overline{QR}\\ \\ \Rightarrow \overline{AR}=x+y\mathrm{....................................................}(\text{iii)}\end{array}$

  $\begin{array}{l}\overline{BP}\text{ = }\overline{BQ}\text{ + }\overline{PQ}\\ \\ \Rightarrow \overline{BP}\text{ = }x\text{ + }y\mathrm{..............................................................}(\text{iv)}\end{array}$

Second Case: $\Delta PQR$ is isosceles triangle. $\overline{PQ}\cong \overline{QR}$

As equilateral $\Delta APQ$ is made on $\overline{PQ}\text{ of }\Delta PQR$ :

  $\overline{AP}\cong \overline{PQ}\cong \overline{QA}\text{ = }y\mathrm{.....................................}(\text{v)}$

Similarly for equilateral $\Delta BQR$ is made on $\overline{QR}\text{ of }\Delta PQR$ : $$

  $\overline{BQ}\cong \overline{QR}\cong \overline{RB}\text{ = y }\mathrm{.................................}\text{(vi)}$

Taking values from (i) and (ii) to find $\overline{AR}\text{ and }\overline{BP}$ :

  $\begin{array}{l}\overline{AR}=\overline{AQ}\text{ + }\overline{QR}\\ \\ \Rightarrow \overline{AR}=y+y=\text{ 2}y\mathrm{................................................}(\text{vii)}\end{array}$

  $\begin{array}{l}\overline{BP}\text{ = }\overline{BQ}\text{ + }\overline{PQ}\\ \\ \Rightarrow \overline{BP}\text{ = }y\text{ + }y=\text{ 2}y\mathrm{.........................................................}(\text{viii)}\end{array}$

Combining results of (iii) and (iv), (vii) and (viii) we get:

  $\overline{AR}\cong \overline{BP}$

Which proves the required result.