Question

7. ABCD is a rhombus. Show that diagonal AC
bisects LA as well as 2 C and diagonal BD
bisects ZB as well as ZD.​

Answer 1

Hyyy....

Given : ABCD is a rhombus, i.e.,

 AB = BC = CD = DA. 

To Prove : ∠DAC = ∠BAC,

∠BCA = ∠DCA 

∠ADB = ∠CDB,

∠ABD = ∠CBD 

Proof : In ∆ABC and ∆CDA, we have 

AB = AD [Sides of a rhombus] 

AC = AC [Common] 

BC = CD [Sides of a rhombus] 

∆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.