PQ = RS terbukti benar, ditunjukkan pada penjelasan di bawah ini.
Misalkan [tex]MN=d[/tex] ([tex]d[/tex] bukan diameter), [tex]MC=r_1[/tex], [tex]NA=r_2[/tex], [tex]\angle AMN=\theta_1[/tex], dan [tex]\angle CNM=\theta_2[/tex].
Maka berlaku:
[tex]\begin{aligned}&\sin\theta_1=\frac{r_2}{d}{\sf\ dan\ }\sin\theta_2=\frac{r_1}{d}.\end{aligned}[/tex]
Sehingga:
[tex]\begin{aligned}&\underline{\begin{array}{l|l}\begin{aligned}PQ&=2r_1\sin\theta_1\\\vphantom{\bigg|}&=2r_1\cdot\frac{r_2}{d}\\\vphantom{\bigg|}PQ&=\frac{2r_1r_2}{d}\\\end{aligned}&\begin{aligned}RS&=2r_2\sin\theta_2\\\vphantom{\bigg|}&=2r_2\cdot\frac{r_1}{d}\\\vphantom{\bigg|}RS&=\frac{2r_1r_2}{d}\\\end{aligned}\end{array}}\\&\quad\therefore\ PQ=RS\quad\blacksquare\end{aligned}[/tex]
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Verified answer
PQ = RS terbukti benar, ditunjukkan pada penjelasan di bawah ini.
Penjelasan dengan langkah-langkah:
Misalkan [tex]MN=d[/tex] ([tex]d[/tex] bukan diameter), [tex]MC=r_1[/tex], [tex]NA=r_2[/tex], [tex]\angle AMN=\theta_1[/tex], dan [tex]\angle CNM=\theta_2[/tex].
Maka berlaku:
[tex]\begin{aligned}&\sin\theta_1=\frac{r_2}{d}{\sf\ dan\ }\sin\theta_2=\frac{r_1}{d}.\end{aligned}[/tex]
Sehingga:
[tex]\begin{aligned}&\underline{\begin{array}{l|l}\begin{aligned}PQ&=2r_1\sin\theta_1\\\vphantom{\bigg|}&=2r_1\cdot\frac{r_2}{d}\\\vphantom{\bigg|}PQ&=\frac{2r_1r_2}{d}\\\end{aligned}&\begin{aligned}RS&=2r_2\sin\theta_2\\\vphantom{\bigg|}&=2r_2\cdot\frac{r_1}{d}\\\vphantom{\bigg|}RS&=\frac{2r_1r_2}{d}\\\end{aligned}\end{array}}\\&\quad\therefore\ PQ=RS\quad\blacksquare\end{aligned}[/tex]