Question Number 191841 by Mingma last updated on 01/May/23
Answered by som(math1967) last updated on 01/May/23
$${from}\bigtriangleup{ADC} \\ $$$$\:\frac{{AD}}{{sin}\mathrm{30}}=\frac{{AC}}{{sin}\mathrm{110}} \\ $$$$\Rightarrow\frac{{AD}}{\mathrm{sin}\:\mathrm{30}}=\frac{{AC}}{\mathrm{sin}\:\mathrm{70}}\:……\mathrm{1} \\ $$$${from}\:\bigtriangleup{ABD} \\ $$$$\:\frac{\mathrm{sin}\:\mathrm{20}}{{AD}}=\frac{\mathrm{sin}\:\mathrm{40}}{{BD}}\:\:\:…..\mathrm{2} \\ $$$$\left(\mathrm{1}\right)×\left(\mathrm{2}\right) \\ $$$$\:\frac{\mathrm{sin}\:\mathrm{20}}{\mathrm{sin}\:\mathrm{30}}=\frac{{AC}}{{BD}}×\frac{\mathrm{sin}\:\mathrm{40}}{\mathrm{sin}\:\mathrm{70}} \\ $$$$\Rightarrow\frac{\mathrm{sin}\:\mathrm{20}×{sin}\mathrm{70}}{\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{sin}\:\mathrm{40}}=\frac{{AC}}{{BD}} \\ $$$$\Rightarrow\frac{\mathrm{2}{sin}\mathrm{20}{cos}\left(\mathrm{90}−\mathrm{70}\right)}{{sin}\mathrm{40}}=\frac{{AC}}{{BD}} \\ $$$$\Rightarrow\frac{{sin}\mathrm{40}}{{sin}\mathrm{40}}=\frac{{AC}}{{BD}} \\ $$$$\therefore{AC}={BD} \\ $$
Answered by HeferH last updated on 04/May/23
Commented by HeferH last updated on 04/May/23
$$\mathrm{AC}=\mathrm{BD} \\ $$