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.
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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.
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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.