Question

Diagonal AC of a parallelogram ABCD bisects ∠A . Show that ABCD is a rhombus

Answer 1

Given

★ ∠1 = ∠2

★ AD = BC

★ AB = DC

To prove

★ ABCD is a rhombus

Solution

⭐$\orange{∠1 = ∠2}$ (given)

⭐ $\orange{∠1 = ∠3}$( alternate angles)

⭐ $\orange{∠2 = ∠4}$(alternate angles)

so ∠1 = ∠2 = ∠3 = ∠4

In ∆ ABC and ∆ADC

⭐ $\purple{AC = AC}$ ( COMMON)

⭐$\purple{∠1 = ∠2}$(GIVEN)

⭐ $\purple{∠3 = ∠4 }$(JUST PROOVED)

∆ABC ≌ ∆ ADC by ASA

⭐ $\red{AB = AD}$ ( by cpct)

⭐$\green{AB = DC}$ (GIVEN) (1)

⭐$\green{AB = AD}$ (CPCT) (2)

⭐$\green{AD = BC}$ (GIVEN).(3)

From equation 1st , 2nd and 3rd

AB = BC = CD = DA

Hence ABCD is a Rhombus

√$\bold\green\star\bold\green{verified} \space\bold\green {answer}$

Answer 2

${\huge{\mathfrak{\pink{\underline{\underline{QuEsTiOn}}}}}}$

Diagonal AC of a parallelogram ABCD bisects ∠A . Show that ABCD is a rhombus

${\huge{\mathfrak{\pink{\underline{\underline{Answer}}}}}}$

${\huge{\underline{\underline{Given:}}}}$

ABCD is a parallelogram..

Diagonal AC bisects angle A

so, angle CAB=angle CAD

${\huge{\underline{\underline{To\ prove:}}}}$

ABCD is a rhombus

means all sides are equal...

AB=BC=CD=DA

${\huge{\underline{\underline{Proof:}}}}$

In triangle ADC and triangle CBA,

AD=CB[Opposite sides are equal of a ||gm]

DC=BA[Opposite sides are equal of a ||gm]

AC=AC[Common]

so,triangle ADC Congruent to triangle CBA[S.S.S. congruence rule]

So,

Angle ACD=AngleCAB[C.P.C.T.]------eq. 1

Angle BCA=AngleCAD[C.P.C.T.]-------eq. 2

Angle CAB=AngleCAD[Given]----------eq. 3

From eq. 1,2 and 3 we see that..

All four angles are equal to each other..

angle ACD=angle CAB=angle BCA=angle CAD

so,it is clear that..

angle ACD=angle CAD

AD=CD[Opposite sides of equal angle of a triangle are equal]

As,AB=CD and AD=BC[opposite side of a parallelogram are equal]

AB=CD=BC=DA

As,all sides are equal now,so it is proved that it is a rhombus..