### Angles in circle, tangent

Question Sample Titled 'Angles in circle, tangent'

The figure shows a circle passing through points ${A}$ , ${B}$ , ${C}$ and ${D}$ such that ${A}{B}={A}{D}$ and ${B}{C}={2}{C}{D}$ . ${A}{T}$ is the tangent to the circle ${A}$ . ${C}{D}$ is produced to meet the tangent at point ${T}$ .

${A}$${B}$${C}$${D}$${T}$

 (a) By considering $\triangle{A}{B}{C}$ and $\triangle{A}{D}{T}$ , prove that ${A}{B}^{{2}}$ $={B}{C}\cdot{D}{T}$ . (3 marks) (b) Prove that ${A}{C}^{{2}}={2}{C}{D}^{{2}}+{A}{D}^{{2}}$ . (3 marks)

 (a) $\angle{A}{B}{C}$ $=\angle{T}{D}{A}$ ext. ∠, cyclic quad. ∵ ${A}{B}$ $={A}{D}$ given ∴ $\angle{B}{C}{A}$ $=\angle{D}{C}{A}$ equal chords, equal ∠s $\angle{D}{C}{A}$ $=\angle{D}{A}{T}$ ∠ in alt. segment ∴ $\angle{B}{C}{A}$ $=\angle{D}{A}{T}$  $\angle{B}{A}{C}$ $={180}^{\circ}-\angle{A}{B}{C}-\angle{B}{C}{A}$ ∠ sum of △  $={180}^{\circ}-\angle{T}{D}{A}-\angle{D}{A}{T}$  $=\angle{D}{T}{A}$ ∠ sum of △ ∴ $\triangle{A}{B}{C}$ ~ $\triangle{T}{D}{A}$ AAA $\dfrac{{{A}{B}}}{{{T}{D}}}$ $=\dfrac{{{B}{C}}}{{{D}{A}}}$ corr. sides, ∼△s ${A}{B}\cdot{D}{A}$ $={B}{C}\cdot{T}{D}$ ∵ ${A}{B}$ $={A}{D}$ ∴ ${A}{B}^{{2}}$ $={B}{C}\cdot{T}{D}$  (3 marks)：Correct proof with reasons (2 marks)：Correct proot with missing reasons (1 marks)：Incomplete proof with any one correct step and one correct reason   (b) Let $\angle{B}{C}{A}$ $=\angle{D}{C}{A}={x}$ . In $\triangle{A}{B}{C}$ , by the cosine formula, ${\cos{{x}}}$ $=\dfrac{{{A}{C}^{{2}}+{B}{C}^{{2}}-{A}{B}^{{2}}}}{{{2}{\left({A}{C}\right)}{\left({B}{C}\right)}}}$ ∵ ${A}{B}$ $={A}{D}$ and ${B}{C}={2}{C}{D}$ ∴ ${\cos{{x}}}$ $=\dfrac{{{A}{C}^{{2}}+{\left({2}{C}{D}\right)}^{{2}}-{A}{D}^{{2}}}}{{{2}{\left({A}{C}\right)}{\left({2}{C}{D}\right)}}}$ ${\cos{{x}}}$ $=\dfrac{{{A}{C}^{{2}}+{4}{C}{D}^{{2}}-{A}{D}^{{2}}}}{{{4}{\left({A}{C}\right)}{\left({C}{D}\right)}}}$ $\ldots{\left({1}\right)}$ In $\triangle{A}{C}{D}$ by the cosine formula ${\cos{{x}}}$ $=\dfrac{{{A}{C}^{{2}}+{C}{D}^{{2}}-{A}{D}^{{2}}}}{{{2}{\left({A}{C}\right)}{\left({C}{D}\right)}}}$ $\ldots{\left({2}\right)}$ From ${\left({1}\right)}$ and ${\left({2}\right)}$ , we have $\dfrac{{{A}{C}^{{2}}+{4}{C}{D}^{{2}}-{A}{D}^{{2}}}}{{{4}{\left({A}{C}\right)}{\left({C}{D}\right)}}}$ $=\dfrac{{{A}{C}^{{2}}+{C}{D}^{{2}}-{A}{D}^{{2}}}}{{{2}{\left({A}{C}\right)}{\left({C}{D}\right)}}}$ ${\left({A}{C}^{{2}}+{4}{C}{D}^{{2}}-{A}{D}^{{2}}\right)}$ $={2}{\left({A}{C}^{{2}}+{C}{D}^{{2}}-{A}{D}^{{2}}\right)}$ ${\left({A}{C}^{{2}}+{4}{C}{D}^{{2}}-{A}{D}^{{2}}\right)}$ $={2}{A}{C}^{{2}}+{2}{C}{D}^{{2}}-{2}{A}{D}^{{2}}$ ${2}{C}{D}^{{2}}+{A}{D}^{{2}}$ $={A}{C}^{{2}}$ ${A}{C}^{{2}}$ $={2}{C}{D}^{{2}}+{A}{D}^{{2}}$  (3 marks)：Correct proof with reasons (2 marks)：Correct proot with missing reasons (1 marks)：Incomplete proof with any one correct step and one correct reason

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