In abc Ab=Ac and D is point on side Ac such that (bc) whole saquare ac.cd prove Bd=Bc
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To prove: BD = BCproof:BC2 = AC {tex} \\times{/tex} CD{tex}\\Rightarrow {/tex}{tex}\\frac{{AC}}{{BC}} = \\frac{{BC}}{{CD}}{/tex} ….(1)Also, {tex}\\angle{/tex} ACB = {tex}\\angle{/tex} BCD ….(2) …..[Common angle]In view of (1) and (2),{tex}\\triangle {/tex} ABC {tex} \\sim {/tex}{tex}\\triangle {/tex}BDC …..[SAS similarity criterion]{tex}\\therefore {/tex}{tex}\\frac{{AC}}{{BC}} = \\frac{{AB}}{{BD}}{/tex} …….({tex}\\because {/tex} corresponding sides of two similar triangles are proportional)But AB = AC …..(Given)BD = BC