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

Answer:

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

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Answer 2

Answer:

Step-by-step explanation: see the solution pic