Higher November 2021 Paper 3 Q23
23 \(ABC\) is a triangle.
\(D\) is the point on \(BC\) such that angle \(BAD\) = angle \(DAC = x^\circ\)
Prove that \(\dfrac{AB}{BD} = \dfrac{AC}{DC}\) (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| Proof (supported) | M1 | for using the sine rule on triangle \(ABD\) or on triangle \(ADC\), to involve sides \(AB\), \(BD\), \(AC\), or \(DC\) eg \(\dfrac{AB}{\sin ADB} = \dfrac{BD}{\sin x}\) oe or \(\dfrac{AC}{\sin ADC} = \dfrac{DC}{\sin x}\) oe OR for an expression for the area of triangle \(ABD\) or for the area of triangle \(ADC\) eg \(\dfrac{1}{2}AB\,AD \sin x\) or \(\dfrac{1}{2}AD\,AC \sin x\) or \(\dfrac{1}{2}h\,BD\) or \(\dfrac{1}{2}h\,DC\) |
| M1 | for using the sine rule on both triangle \(ABD\) and on triangle \(ADC\), to involve sides \(AB\), \(BD\), \(AC\), or \(DC\) eg \(\dfrac{AB}{\sin ADB} = \dfrac{BD}{\sin x}\) oe and \(\dfrac{AC}{\sin ADC} = \dfrac{DC}{\sin x}\) oe OR for two expressions for the area of either triangle \(ABD\) or for triangle \(ADC\) eg \(\dfrac{1}{2}AB\,AD \sin x\) and \(\dfrac{1}{2}h\,BD\) or \(\dfrac{1}{2}AD\,AC \sin x\) and \(\dfrac{1}{2}h\,DC\) | |
| M1 | for stating or showing \(\sin ADB = \sin ADC\), eg \(\sin y = \sin(180 - y)\) OR for using two expressions to form an equation eg \(\dfrac{\frac{1}{2}AB\,AD \sin x}{\frac{1}{2}AD\,AC \sin x} = \dfrac{\frac{1}{2}h\,BD}{\frac{1}{2}h\,DC}\) oe | |
| C1 | for a full method to arrive at the given answer |
Additional guidance
Accept extra letters eg \(y\) shown on diagram for any angle used