In a triangle ABC, prove $\frac{a}{{{\text{sin}}\,A}} = \frac{b}{{{\text{sin}}\,B}} = \frac{c}{{{\text{sin}}\,C}}$.

Prove that the area of the triangle ABC is $\frac{1}{2}ab\,{\text{sin}}\,{\text{C}}$.

Given that R denotes the radius of the circumscribed circle prove that $\frac{a}{{{\text{sin}}\,A}} = \frac{b}{{{\text{sin}}\,B}} = \frac{c}{{{\text{sin}}\,C}} = 2R$.

Hence show that the area of the triangle ABC is $\frac{{abc}}{{4R}}$.

Find in terms of R, the two values of (DE)² such that the area of the shaded region is twice the area of the triangle DEF.

Using two diagrams, explain why there are two values of (DE)².

A parallelogram is positioned inside a circle such that all four vertices lie on the circle. Prove that it is a rectangle.

${\text{sin}}\,{\text{B}} = \frac{h}{c}$ and ${\text{sin}}\,{\text{C}} = \frac{h}{b}$      M1A1

hence $h = c\,{\text{sin}}\,{\text{B}} = b\,{\text{sin}}\,{\text{C}}$      A1

by dropping a perpendicular from B, in exactly the same way we find $c\,{\text{sin}}\,{\text{A}} = a\,{\text{sin}}\,{\text{C}}$      R1

hence $\frac{a}{{{\text{sin}}\,A}} = \frac{b}{{{\text{sin}}\,B}} = \frac{c}{{{\text{sin}}\,C}}$

[4 marks]

area = $\frac{1}{2}ah$      M1A1

= $\frac{1}{2}ab\,{\text{sin}}\,{\text{C}}$      AG

[2 marks]

since the angle at the centre of circle is twice the angle at the circumference ${\text{sin}}\,A = \frac{a}{{2R}}$         M1A1

hence $\frac{a}{{{\text{sin}}\,A}} = 2R$ and therefore $\frac{a}{{{\text{sin}}\,A}} = \frac{b}{{{\text{sin}}\,B}} = \frac{c}{{{\text{sin}}\,C}} = 2R$      AG

[2 marks]

area of the triangle is $\frac{1}{2}ab\,{\text{sin}}\,{\text{C}}$      M1

since ${\text{sin}}\,{\text{C}} = \frac{c}{{2R}}$        A1

area of the triangle is $\frac{1}{2}ab\,\frac{c}{{2R}} = \frac{{abc}}{{4R}}$      AG

[2 marks]

area of the triangle is $\frac{{\pi {R^2}}}{6}$      (M1)A1

(DE)² + (EF)² = 4R²       M1

(DE)² = 4R² −  (EF)² 

$\frac{1}{2}\left( {{\text{DE}}} \right)\left( {{\text{EF}}} \right) = \frac{{\pi {R^2}}}{6} \Rightarrow \left( {{\text{EF}}} \right) = \frac{{\pi {R^2}}}{{3\left( {{\text{DE}}} \right)}}$      M1A1

${\left( {{\text{DE}}} \right)^2} = 4{R^2} - \frac{{{\pi ^2}{R^4}}}{{9{{\left( {{\text{DE}}} \right)}^2}}}$      A1

$9{\left( {{\text{DE}}} \right)^4} - 36{\left( {{\text{DE}}} \right)^2}{R^2} + {\pi ^2}{R^4} = 0$      A1 

${\left( {{\text{DE}}} \right)^2} = \frac{{36{R^2} \pm \sqrt {1296{R^4} - 36{\pi ^2}{R^4}} }}{{18}}$     M1

${\left( {{\text{DE}}} \right)^2} = \frac{{36{R^2} \pm 6{R^2}\sqrt {36 - {\pi ^2}} }}{{18}}\left( { = \frac{{6{R^2} \pm {R^2}\sqrt {36 - {\pi ^2}} }}{3}} \right)$      A1

[9 marks]

      A1A1

[2 marks]

$\mathop {\text{A}}\limits^ \wedge   + \mathop {\text{C}}\limits^ \wedge   = 180^\circ$ (cyclic quadrilateral)      R1

however $\mathop {\text{A}}\limits^ \wedge   = \mathop {\text{C}}\limits^ \wedge$ (ABCD is a parallelogram)       R1

$\mathop {\text{A}}\limits^ \wedge   = \mathop {\text{C}}\limits^ \wedge   = 90^\circ$        A1

$\mathop {\text{B}}\limits^ \wedge   = \mathop {\text{D}}\limits^ \wedge   = 90^\circ$

hence ABCD is a rectangle        AG

[3 marks]