If one diagonal of a Trapezium …

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Sia ? 6 years, 5 months ago
DE = EB = 1:3
In {tex}\triangle{/tex}AEB and {tex}\triangle{/tex}CED, {tex}\angle 1 = \angle 2{/tex} (alternate angles)
{tex}\angle 3 = \angle 4{/tex} (Vertically opposite angles.)
{tex}\therefore \quad \Delta \mathrm { AEB } \sim \Delta \mathrm { CED }{/tex}
{tex}\Rightarrow \quad \frac { \mathrm { AB } } { \mathrm { CD } } = \frac { \mathrm { BE } } { \mathrm { DE } } \Rightarrow \frac { \mathrm { AB } } { \mathrm { CD } } = \frac { 3 } { 1 }{/tex} [{tex}\because {/tex} DE: BE = 1:3]
{tex}\Rightarrow{/tex} AB = 3CD
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