Rectangle Maths Tuition

Solution:

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

$\angle PCS=\angle SQR$ (given)

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

(b)(I)

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

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

Hence $\triangle KQR$ is an isosceles triangle.

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

Hence,

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

(shown)

(ii) By similar triangles,

$\displaystyle\frac{PC}{PS}=\frac{RQ}{RS}=\frac{PS}{PQ}=\frac{PS}{2PS}=\frac{1}{2}$

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

Hence

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

(shown)

Author: mathtuition88