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Answer :
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.