Question

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

Answer 1

Step-by-step explanation:

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MATHS

Diagonal AC of a parallelogram ABCD bisects ∠A

Show that:

(i) it bisects ∠C also,

(ii) ABCD is a rhombus

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ANSWER

(i) Here, ABCD is a parallelogram and diagonal AC bisects ∠A.

∴ ∠DAC=∠BAC ---- ( 1 )

Now,

AB∥DC and AC as traversal,

∴ ∠BAC=∠DCA [ Alternate angles ] --- ( 2 )

AD∥BC and AAC as traversal,

∴ ∠DAC=∠BCA [ Alternate angles ] --- ( 3 )

From ( 1 ), ( 2 ) and ( 3 )

∠DAC=∠BAC=∠DCA=∠BCA

∴ ∠DCA=∠BCA

Hence, AC bisects ∠C.

(ii) In △ABC,

⇒ ∠BAC=∠BCA [ Proved in above ]

⇒ BC=AB [ Sides opposite to equal angles are equal ] --- ( 1 )

⇒ Also, AB=CD and AD=BC [ Opposite sides of parallelogram are equal ] ---- ( 2 )

From ( 1 ) and ( 2 ),

⇒ AB=BC=CD=DA

Hence, ABCD is a rhombus

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..