ABCD is a rhombus show that …
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Preeti Dabral 1 year, 6 months ago
Given: ABCD is a rhombus
In {tex}\triangle{/tex}ABC and {tex}\triangle{/tex}ADC,
AB = CD [Sides of a rhombus]
BC = DA [Sides of a rhombus]
AC = AC [Common]
{tex}\therefore{/tex} {tex}\triangle{/tex}ABC {tex}\cong{/tex} {tex}\triangle{/tex}ADC [By SSS Congruency]
{tex}\therefore{/tex} {tex}\angle {/tex}CAB = {tex}\angle {/tex}CAD And {tex}\angle {/tex}ACB = {tex}\angle {/tex}ACD
Hence AC bisects {tex}\angle {/tex}A as well as {tex}\angle {/tex}C
Similarly, by joining B to D, we can prove that {tex}\triangle{/tex}ABD {tex}\cong{/tex} {tex}\triangle{/tex}CBD
Hence BD bisects {tex}\angle {/tex}B as well as {tex}\angle {/tex}D
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