Chapter 5.6, Problem 3PSA

To determine

To prove: Quadrilateral SMPR is a parallelogram.

Explanation of Solution

Given information:The end points of the quadrilateral SMPRare at the circumference of $\odot O$ with its diagonals passing through the center of given circle.

Concept used:

(a) Side-Angle-Side (SAS) Postulate of Congruency of triangles.

(b) If both pair of opposite sides are parallel to each other then the quadrilateral is a parallelogram.

Proof:We can clearly see it in the given diagram that the diagonals SPand RMare also the diameter of $\odot O$ then,

As per Thales’ theorem, “Inscribed angles subtended by a diameter are always right”

So, $\angle SMP=\angle RPM={90}^{\circ }$ ……………………. (i)

Now, in $△SMP$ and $△RPM$

  $SP=RM$ (As SP and RMare the diameters of $\odot O$ )

  $\angle SMP=\angle RPM$ (from eq(i))

  $MP=MP$ (Side common to both the triangles)

By Side-Angle-Side (SAS) Postulate of Congruency of triangles we have,

  $△SMP\cong △RPM$

Then, $SM=RP$ (Corresponding Parts of Congruent Triangles)

  $\angle MSP=\angle SPR$ (Corresponding Parts of Congruent Triangles)

Also, $\angle MSP$ and $\angle SPR$ are alternate interior angles for lines SMand RP where SP can be its traversal intersecting these two lines, these alternate interior angles are equal as proved above which means,

  $SM\parallel RP$

Thus, $SM=RP$ and $SM\parallel RP$

Hence, a pair of opposite sides is equal and parallel to each other so the given quadrilateral SMPRis a parallelogram.