O Level 2007 E Maths Paper 2 Q3 Solution (Geometry Question)

Posted on June 2, 2013 by

(a)
\angle DAB=\angle DCB
(opposite angles of parallelogram)

\displaystyle\angle PAD=\frac{1}{2}\angle DAB=\frac{1}{2}\angle DCB=\angle RCB
(shown)

(b)
AD=BC

\angle PAD=\angle RCB
(from part a)

\angle ADC=\angle ABC
(opposite angles of parallelogram)

\displaystyle\angle ADP=\frac{1}{2}\angle ADC=\frac{1}{2} ABC=\angle RBC

Thus, triangles ADP and CBR are congruent (ASA).

(c)(i)
\angle ADC+\angle DAB=180^\circ
\angle DAP+\angle ADP=90^\circ

Considering the triangle ADP,
\angle DPA=180^\circ-90^\circ=90^\circ
(shown)

(ii)
\angle DAB+\angle ABC=180^\circ

\angle BAQ+\angle ABQ=90^\circ

Considering the triangle ABQ,
\angle PQR=180^\circ-90^\circ=90^\circ
(shown)

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