\(ABCD\) is a parallelogram and \(AP\) and \(CQ\) are perpendiculars from vertices \(A\) and \(C\) on diagonal \(BD\). Show that
(i) \(∆APB ≅ ∆CQD\) and
(ii) \(AP = CQ\)
Proof:
\(ABCD\) is a parallelogram
(i) In \(∆APB\) and \(∆CQD\), we have
(i) In \(∆APB\) and \(∆CQD\), we have
\(∠APB = ∠CQD=\)\(^{\circ}\)
\(AB = CD\) [Opposite sides of the parallelogram are ]
As \(AB || CD\) and \(BD\) is a transversal, \(∠ABP = ∠CDQ\) [Alternate angles are equal]
Hence, \(∆APB ≅ ∆CQD\) [By ]
(ii) As, \(∆APB ≅ ∆CQD\)
\(AP = CQ\) [By ]
Hence, proved.