ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that
(i) \(\triangle{APB}\) \(\displaystyle \cong \) \(\triangle{CQD}\)
(ii) AP = CQ

Given: ABCD is a parallelogram and AP and CQ are perpendicular from vertices A and C on diagonal BD.

\(\because \) AB || CD and BD is a transversal, we get,

\(\angle{CDB}\) = \(\angle{DBA}\) ...(i)
Now, in \(\triangle{APB}\) and \(\triangle{CQD}\), we have,
CD = AB ...(Sides of parallelogram)
\(\angle{CQD}\) = \(\angle{APB}\) = \(90^\circ\) ...(Given)
\(\angle{CDQ}\) = \(\angle{ABP}\) ...(From Eq. (i))
\(\therefore \) \(\triangle{APB}\) \(\displaystyle \cong \) \(\triangle{CQD}\) ...(By ASA rule)
\(\therefore \) AP = CQ ...(By CPCT)
Hence, it is proved.