Show that the bisectors of the angles of a parallelogram enclose a rectangle.

Explanation

Since DC||AB and DA cuts them, we have
`angle A + angle D = 180^(@) " " ("co-interior " angle)`
`rArr (1)/(2) angle A + (1)/(2) angle D = 90^(@)`
`rArr angle PAD + angle PDA=90^(@)`
`rArr angle APD=90^(@) " "[ because " sum of the " angle " of a " triangle " is " 180^(@)]`
`rArr angle SPQ =90^(@) " " ("vert. opp. " angle ).`
Now, AD||BC and DC cuts them.
`therefore angle D + angle C = 180^(@) " "("co-interior " angle )`
`rArr (1)/(2) angle D +(1)/(2) angle C =90^(@)`
`rArr angle QDC + angle QCD =90^(@)`
`rArr angle DQC =90^(@) " "[ because " sum of the " angle " of a " triangle " is " 180^(@)]`
`rArr angle PQR =90^(@).`
Similarly, ` angle QRS =90^(@) and angle RSP =90^(@)`.
Thus, PQRS is quadrilateral each of whose angles is `90^(@).`
Hence, PQRS is a rectangle.