Collinear Proof in Circle Geometry

(a) \( \ \ \) Show that \( \angle DSR = \angle DAR \).

The quadrilateral \( DRAS \) is cyclic, since \( \angle DRA + \angle DSA = \pi (180^{\circ}) \).
Therefore \( \angle DSR = \angle DAR \) (angles in the same segment of circle \( DRAS \), both standing on same chord \( DR \).

(b) \( \ \ \) Show that \( \angle DST = \pi – \angle DCT \).

Since \( \angle DSC = \angle DTC \) both \( \displaystyle \frac{\pi}{2} \) and these stand on the same interval \( DC \) and on the same side of it, then \( DSTC \) is a cyclic quadrilateral.
Thus \( \angle DST + \angle DCT = \pi \) (opposite angles of cyclic quadrilateral \( DSTC \))
\( \therefore \angle DST = \pi – \angle DCT \)

(c) \( \ \ \) Deduce that the points \( R, S \) and \( T \) are collinear.

\( \angle DSR + \angle DST = \angle DAR + \pi – \angle DCT \cdots (1) \)
\( \angle DAR = \angle DCB \) (exterior angle of cyclic quadrilateral \( DABC \) equals the interior opposite angle).
\( \angle DCB = \angle DCT \) (same angle)
\( \angle DAR = \angle DCT \)
\( \angle DSR + \angle DST = \angle DCT + \pi – \angle DCT \cdots (1) \)
\( \angle DSR + \angle DST = \pi \)
Thus \( \angle RST \) is a straight angle, and therefore points \( R,S \) and \( T \) are collinear.

Algebra Algebraic Fractions Arc Binomial Expansion Capacity Common Difference Common Ratio Differentiation Divisibility Proof Double-Angle Formula Equation Exponent Exponential Function Factorials Factorise Functions Geometric Sequence Geometric Series Index Laws Inequality Integration Kinematics Length Conversion Logarithm Logarithmic Functions Mass Conversion Mathematical Induction Measurement Perfect Square Perimeter Prime Factorisation Probability Proof Pythagoras Theorem Quadratic Quadratic Factorise Rational Functions Sequence Sketching Graphs Surds Time Transformation Trigonometric Functions Trigonometric Properties Volume




Your email address will not be published. Required fields are marked *