Question

In the given diagram ABCD is a parallelogram. ∆APD and ∆BQC are equilateral triangles. Prove that :
(i) ∠PAB = ∠QCD
(ii) PB = QD

Answer 1

1.The given diagram ABCD is a parallelogram

Given,

= ∆PAD and ∆BQC are equilateral

= ∆PAD=∆BCQ=60

= ∆BAD=∆BCD

adding equation

= ∆PAD+∆BAD=∆BCQ+∆BCD

= ∆PAB=∆QCD

( proved)

2.Now in ∆PAB and ∆QCD

= PA=AD

= AD=BC

= BC=QC

= PA=QC

= AB-CD

= ∆PAB=∆QCD

= ∆PAB=∆QCD

= PB=QD

(proved)

Answer 2

Hence Proved.

Explanation:

Given,

$ABCD$ parallelogram, $\triangle APD$ and $\triangle BQC$ are equilateral triangles.

Prove that:

1. $\angle PAB=\angle QCD$
2. $PB=QD$

∵ $ABCD$ parallelogram.

So, Opposite angle are also equal.

∴$\angle BAD=\angle BCD$__1

∵ $\triangle APD$ and $\triangle BQC$ are equilateral triangles.

So, $\angle PAD=\angle QCB=60\ degree$__2

By adding equation-1 and 2,

$\angle BAD+\angle PAD=\angle BCD+\angle QCB$

⇒$\angle PAB=\angle QCD$ (Prove-1)

In $\triangle PAB$ and $\triangle QCD$,

• $AB=CD$ (∵ Opposite sides of $ABCD$ parallelogram)

• $PA=QC$ (∵ $AD=BC$ and $AD=PA$ also $BC=QC$)

• $\angle PAB=\angle QCD$

∴ $\triangle PAB$≅$\triangle QCD$

∴ $PB=QD$ (Prove-2)

Hence Proved.