Question

prove that BD bisects angle B as well as angle D. ​

Answer 1

$\huge{\sf{\underline{\underline{Given}}}}$

ABCD is a rhombus, i.e.,

AB = BC = CD = DA.

$\huge{\sf{\underline{\underline{To ~Prove}}}}$

∠DAC = ∠BAC,

∠BCA = ∠DCA

∠ADB = ∠CDB, ∠ABD = ∠CBD

$\implies\large\bold{Proof}$

In ∆ABC and ∆CDA, we have

$\looparrowright\sf\red{AB = AD (Sides~ of~ a ~rhombus~)}$

$\looparrowright\sf\orange{AC = AC (Common)}$

$\looparrowright\sf\green{BC = CD (Sides~ of ~a ~rhombus)}$

$\looparrowright\sf\pink{∆ABC = ∆ADC (SSS ~congruence) }$

So, ∠DAC = ∠BAC

∠BCA = ∠DCA

$\sf\color{blue}{Similarly, ∠ADB = ∠CDB~ and~ ∠ABD = ∠CBD. }$

$\looparrowright$Hence, diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D. Proved..

Answer 2

Answer:

$\huge\sf\underline\blue{Answer}$

In ∆ABC and ∆CDA, we have

AB = AD (Sides of a rhombus)

↬AB=AD(Sides of a rhombus )

AC = AC (Common)

↬AC=AC(Common)

BC = CD (Sides of a rhombus)

↬BC=CD(Sides of a rhombus)

∆ABC = ∆ADC (SSS congruence)

↬∆ABC=∆ADC(SSS congruence)

So, ∠DAC = ∠BAC

∠BCA = ∠DCA

Similarly, ∠ADB = ∠CDB and ∠ABD = ∠CBD.