Let $A B C D$ be a parallelogram. Let $B P$ and $D Q$ be perpendiculars respectively from $B$ and $D$ on to $A C$ . Prove that $B P = D Q$ .

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Solution

It is given that in parallelogram $A B C D$ , $B P$ is perpendicular to $A C$ and $D Q$ is perpendicular to $A C$ .

In $Δ A D Q$ and $Δ C B P$ ,

$A D = C D$ (Opposite sides of a parallelogram)

$∠ D A Q = ∠ B C P$ and $A D | | B C, A C$ (Transversal alternate angles)

$∠ D Q A = ∠ B P C = 90^{0}$ (Given)

$⸫ Δ A D Q = Δ C B P$ (SAA)

$⸫ B P = D Q$ (C.P.C.T)

Hence proved.

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Let ABCD be a parallelogram. Let BP and DQ be perpendiculars respectively from B and D on to AC. Prove that BP=DQ.

It is given that in parallelogram ABCD, BP is perpendicular to AC and DQ is perpendicular to AC. In ΔADQ and ΔCBP,AD=CD (Opposite sides of a parallelogram)∠DAQ=∠BCP and AD||BC,AC (Transversal alternate angles)∠DQA=∠BPC=900 (Given)⸫ΔADQ=ΔCBP (SAA)⸫BP=DQ (C.P.C.T)Hence proved.

Let $A B C D$ be a parallelogram. Let $B P$ and $D Q$ be perpendiculars respectively from $B$ and $D$ on to $A C$ . Prove that $B P = D Q$ .