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In the diagram ABCD is a cyclic quadrilateral - HSC - SSCE Mathematics Extension 2 - Question 7 - 2010 - Paper 1
Question 7
In the diagram ABCD is a cyclic quadrilateral. The point K is on AC such that \( \angle ADK = \angle LCB \), and hence \( \triangle ADK \) is similar to \( \triangle... show full transcript
Worked Solution & Example Answer:In the diagram ABCD is a cyclic quadrilateral - HSC - SSCE Mathematics Extension 2 - Question 7 - 2010 - Paper 1
Step 1
Show that \( \triangle ADB \) is similar to \( \triangle KDC \)
Answer
To show that ( \triangle ADB ) is similar to ( \triangle KDC ), we need to demonstrate that two angles in one triangle are equal to two corresponding angles in the other triangle. Since both triangles share angle ( \angle ADB ), and we know that ( \angle ADK = \angle LCB ) from the properties of cyclic quadrilaterals, therefore, by the AA criterion for triangle similarity, it follows that ( \triangle ADB \sim \triangle KDC ).
Step 2
Using the fact that \( AC = AK + KC \), show that \( BD \times AC = AD \times BC + AB \times DC \)
Answer
From the similarity established earlier, we can write the corresponding ratios:
[ \frac{AD}{AB} = \frac{KC}{DC} \quad \text{and} \quad \frac{BD}{AC} = \frac{AB}{BC}. ]
Using the first ratio, we can express ( AD ) as ( AD = \frac{AB \cdot KC}{DC} ). Substituting this into the identity for the products, we rearrange to obtain:
[ BD \cdot AC = AD \cdot BC + AB \cdot DC. ]
Thus confirming the required equation.
Step 3
Use the result in part (ii) to show that \( x = \frac{1 + \sqrt{5}}{2} \)
Answer
In a regular pentagon inscribed in a circle, the diagonals can be represented as chord lengths. Let's denote the chord lengths in terms of the angle subtended at the center. Utilizing the properties of symmetry and the derived formula from part (ii), we substitute the side lengths and relationships corresponding with the diagonals.
Given that each side of the pentagon is of unit length, we can set up a geometric relationship that ultimately leads us to:
[ x = \frac{1 + \sqrt{5}}{2}. ]
This finalizes the proof utilizing the given derived relationships.