(i)<\/strong> Since, AC \\(\\perp\\) BD, therefore,<\/p>\n
\\(\\triangle\\) ABC ~ \\(\\triangle\\) DBA\u00a0 and\u00a0 each triangle is similar to whole triangle ABD.<\/p>\n
\\(\\implies\\)\u00a0 \\(AB\\over BD\\) = \\(BC\\over AB\\)<\/p>\n
So, \\({AB}^2\\) = BC.BD<\/strong><\/p>\n
(ii)<\/strong> Since, \\(\\triangle\\) ABC ~ \\(\\triangle\\) DAC, therefore,<\/p>\n
\\(\\implies\\)\u00a0 \\(AC\\over BC\\) = \\(DC\\over AC\\)<\/p>\n
So,\u00a0\\({AC}^2\\) = BC.DC<\/strong><\/p>\n
(iii)<\/strong> Since, \\(\\triangle\\) ACD ~ \\(\\triangle\\) BAD, therefore,<\/p>\n
\\(\\implies\\)\u00a0 \\(AD\\over CD\\) = \\(BD\\over AD\\)<\/p>\n
So,\u00a0\\({AD}^2\\) = BD.CD<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"
Solution : (i) Since, AC \\(\\perp\\) BD, therefore, \\(\\triangle\\) ABC ~ \\(\\triangle\\) DBA\u00a0 and\u00a0 each triangle is similar to whole triangle ABD. \\(\\implies\\)\u00a0 \\(AB\\over BD\\) = \\(BC\\over AB\\) So, \\({AB}^2\\) = BC.BD (ii) Since, \\(\\triangle\\) ABC ~ \\(\\triangle\\) DAC, therefore, \\(\\implies\\)\u00a0 \\(AC\\over BC\\) = \\(DC\\over AC\\) So,\u00a0\\({AC}^2\\) = BC.DC (iii) Since, \\(\\triangle\\) ACD ~ \\(\\triangle\\) …<\/p>\n