Rectangle Maths Tuition



(a) \angle SPC=\angle SRQ=90^\circ

\angle PCS=\angle SQR (given)

Thus \triangle PCS is similar to \triangle RQS (AAA)


\angle KRQ=90^\circ /2=45^\circ (KR bisects \angle QRS)

Thus \angle QKR=180^\circ-90^\circ-45^\circ=45^\circ

Hence \triangle KQR is an isosceles triangle.

So KQ=QR=PS=\frac{1}{2}PQ


\begin{array}{rcl}PK&=&PQ-KQ\\    &=&PQ-\frac{1}{2}PQ\\    &=&\frac{1}{2}PQ\\    &=&\frac{1}{2}(2PS)\\    &=&PS    \end{array}


(ii) By similar triangles,


Thus PC=\frac{1}{2}PS=\frac{1}{2} PK


\begin{array}{rcl}CK&=&PK-PC\\    &=&PK-\frac{1}{2}PK\\    &=&\frac{1}{2}PK\\    &=&PC    \end{array}


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