Question Number 44103 by Raj Singh last updated on 21/Sep/18
Commented by Raj Singh last updated on 21/Sep/18
$${in}\:{this}\:{figure}\:{PQ}\mid\mid{BA}\:{and}\:{PR}\mid\mid{CA} \\ $$$${if}\:{PD}=\mathrm{12}\:{find}\:{BD}×{CD} \\ $$
Answered by MrW3 last updated on 21/Sep/18
$$\Delta{DBR}\sim\Delta{DPQ} \\ $$$$\frac{{BD}}{{PD}}=\frac{{BR}}{{PQ}} \\ $$$$\Delta{DPR}\sim\Delta{DCQ} \\ $$$$\frac{{PD}}{{CD}}=\frac{{PR}}{{CQ}} \\ $$$$\Delta{BPR}\sim\Delta{PCQ} \\ $$$$\frac{{BR}}{{PQ}}=\frac{{PR}}{{CQ}} \\ $$$$\Rightarrow\frac{{BD}}{{PD}}=\frac{{PD}}{{CD}} \\ $$$$\Rightarrow{BD}×{CD}={PD}^{\mathrm{2}} =\mathrm{12}^{\mathrm{2}} =\mathrm{144} \\ $$