Parallelogram Maths Tuition: Solution



(a) We have \angle APQ=\angle ARQ (opp. angles of parallelogram)

AP=RQ (opp. sides of parallelogram)

AR=PQ (opp. sides of parallelogram)

Thus, \triangle APQ\equiv\triangle QRA (SAS)

Similarly, \triangle ABC\equiv\triangle CDA (SAS)

\triangle CHQ\equiv\triangle QKC (SAS)

Thus, \begin{array}{rcl}\text{area of BPHC}&=&\triangle APQ-\triangle ABC-\triangle CHQ\\    &=&\triangle QRA-\triangle CDA-\triangle QKC\\    &=& \text{area of DCKR}    \end{array}



\angle ACD=\angle HCQ (vert. opp. angles)

\angle ADC=\angle CHQ (alt. angles)

\angle DAC=\angle CQH (alt. angles)

Thus, \triangle ADC is similar to \triangle QHC (AAA)

Hence, \displaystyle\frac{AC}{DC}=\frac{QC}{HC}

Thus, AC\cdot HC=DC\cdot QC


Author: mathtuition88

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