Question

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

Answer 1

Step-by-step explanation:

Pls check if it is correct or not

Answer 2

Answer:

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

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..