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Question Number 191841 by Mingma last updated on 01/May/23

	Answered by som(math1967) last updated on 01/May/23

	$f r o m △ A D C$ $\frac{A D}{s i n 30} = \frac{A C}{s i n 110}$ $\Rightarrow \frac{A D}{\sin 30} = \frac{A C}{\sin 70} . . . . . . 1$ $f r o m △ A B D$ $\frac{\sin 20}{A D} = \frac{\sin 40}{B D} . . . . . 2$ $(1) \times (2)$ $\frac{\sin 20}{\sin 30} = \frac{A C}{B D} \times \frac{\sin 40}{\sin 70}$ $\Rightarrow \frac{\sin 20 \times s i n 70}{\frac{1}{2} \times \sin 40} = \frac{A C}{B D}$ $\Rightarrow \frac{2 s i n 20 c o s (90 - 70)}{s i n 40} = \frac{A C}{B D}$ $\Rightarrow \frac{s i n 40}{s i n 40} = \frac{A C}{B D}$ $∴ A C = B D$
	Answered by HeferH last updated on 04/May/23

	Commented by HeferH last updated on 04/May/23

	$AC = BD$
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