Question

ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Answer 1

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In rhombus all side are equal and diagonols are different but intersect at 90°

⇒AB=BC=CD=DA

⇒AC⊥BD

In ΔABC

⇒AB=BC

⇒∴∠CAB=∠ACB

⇒AO=OC

∴∠ABO=∠CBO

and similarly

In ΔADC

⇒AD=DC

⇒AO=OC

∴ corresponding angles are equal

⇒∠DAC=∠DCA and ∠CDO=∠ADO

Hence,AC bisect ∠A and ∠C and BD bisects ∠B and ∠D

Answer 2

Given :-

ABCD is a rhombus.

To Show :-

AC bisect ∠A and ∠C

BD bisect ∠B and ∠D

Solution and concept :-

To show that Diagonal bisect the angles of rhombus , at first we have to draw a rhombus then add Vertex A to C and Vertex B to D such that AC and BD become the bisector of angles of rhombus. As we know that in rhombus all sides are equal and it's opposite sides are also equal. To show diagonal bisect its angles , at first we have to prove congruency as per the figure. Concept, remember bisector always the whole angles bisect into two equal parts:

As per the construction figures :-

∆ADC and ∆BAC

⟼ AD = BC ( AD||BC)

⟼ DC = AB ( DC||AB)

⟼ ∠D = ∠B ( opposite angles are equal)

∆ADC ≅ ∆BAC (by S.A.S criterion of congruency)

⟼ ∠ABC = ∠ABD + ∠CBD ( by CPCTC)

⟼ ∠ABD = ∠CBD ( by CPCTC)

∆DCB and ∆BAD

⟼ AD = BC ( AD||BC)

⟼ DC = AB ( DC||AB)

⟼ ∠A = ∠C ( opposite angles are equal)

∆DCB ≅ ∆BAD (by S.A.S criterion of congruency)

⟼ ∠DAB = ∠DAC + ∠BAC (by CPCTC)

⟼ ∠DAC = ∠BAC ( by CPCTC)

From above statement :-

AD and BD Bisect angles

Hence, proved