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

$\large\bold{Proof}$

In ∆ABC and ∆CDA, we have

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

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

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

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

So, ∠DAC = ∠BAC

∠BCA = ∠DCA

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

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

Answer 2

Step-by-step explanation:

∠DAC = ∠BAC,

∠BCA = ∠DCA

∠ADB = ∠CDB, ∠ABD = ∠CBD

$\large\bold{Proof}$

In ∆ABC and ∆CDA, we have

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

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

So, ∠DAC = ∠BAC

∠BCA = ∠DCA

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

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