Chapter 9.5, Problem 18WE

To determine

To name two pairs to similar triangles such that each triangle has a vertex $V$ .

Answer to Problem 18WE

  $\Delta VPR\approx \Delta VSQ$ and $\Delta VPS\approx \Delta VQR$

Explanation of Solution

Given:

Inscribed quadrilateral $PQRS$ with shortest side as $PS$ .

Diagonals intersect at $T$ .

  $QP$ and $RS$ extended to meet at $V.$

  

In $\Delta VPR$ and $\Delta VSQ$ ,

  $\angle V$ is common…(i)

As $\angle PQS$ and $\angle PRS$ form the same arc $\stackrel{⌢}{PS}$ within the circle, thus they are equal by the property of circle.

  $\angle PQS=\angle PRS$

Which also means $\angle VQS=\angle PRV$ …(ii)

Thus, from equation (i) and (ii)and by angle-angle theorem of similarity of triangles

  $\Rightarrow \Delta VPR\approx \Delta VSQ$

Now in $\Delta VPS$ and $\Delta VQR$ ,

  $\angle V$ is common…(iii)

Now as segment QP and segment RS extend and meet at a common point V outside the circle, while passing through the ends of segment PS , it can be said that $PS\parallel QR$ .

As a result, $\angle VPS=\angle VQR$ …(iv) as they’re alternate angles formed by intersection of segment QV with parallel lines.

Thus, from equation (iii) and (iv)and by angle-angle theorem of similarity of triangles

  $\Rightarrow \Delta VPS\approx \Delta VQR$

Therefore, the two pair of similar triangles are $\Delta VPR\cong \Delta VSQ$ and $\Delta VPS\cong \Delta VQR$ .

Conclusion:

Therefore, the two pair of similar triangles are $\Delta VPR\cong \Delta VSQ$ and $\Delta VPS\cong \Delta VQR$ this can easily be proved by using angle-angle theorem of similarity of triangles.