Question

PQRS is a parallelogram in which X is the mid-point of PQ and SX bisects angle PSR. Prove that [i] RX bisects angle SRQ ​[ii] angle SXR = 90 degrees

Answer 1

in pllgm PQRS 

<PSRS+<QRS =180  (ADJACENT ANGLES OF A PLLGM ARE                                                                                                SUPPLEMENTARY)
SO,1/2<PSR=,XSR
                               (AS SX IS BISECTOR OF PSR)
SO SIMILARLY,1/2<QRS=<XRS
                                    THUS RX IS BISECTOR OF <QRS...PROVED (i)

AS,1/2<PSR+1/2<QRS=<XSR+<XRS(SHOWN ABOVE)
            =90'=<XSR+<XRS
SO, AS <XSR AND <XRS ARE BISECTORS THUS BOTH ARE 45"

SO, IN TRI.XSR 
                              =<XSR+<XRS+<SXR=180
                              =<SXR=180-90
                                          =90'
                                                      HENCE PROVED <SXR=90'
#shimmers