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Question Number 95177 by mr W last updated on 23/May/20

Commented by mr W last updated on 23/May/20

$$A,B,C,Darecentersofthesquares.$$$$ProvethatAC=BDandAC\mathrm{\perp }BD.$$

Commented by PRITHWISH SEN 2 last updated on 24/May/20

$$\mathrm{sir}\mathrm{I}\mathrm{have}\mathrm{a}{\mathrm{sol}}^{\mathrm{n}}\mathrm{with}\mathrm{the}\mathrm{help}\mathrm{of}\mathrm{complex}\mathrm{number}$$$$\left(\mathrm{though}\mathrm{it}\mathrm{is}\mathrm{not}\mathrm{mine}\right).$$

Commented by mr W last updated on 24/May/20

$$pleaseshareit!thanks!$$$$allmethodsarewelcome.$$

Commented by PRITHWISH SEN 2 last updated on 24/May/20

$$\mathrm{OK}\mathrm{sir}\mathrm{I}\mathrm{will}\mathrm{post}\mathrm{it}\mathrm{later}$$

Commented by PRITHWISH SEN 2 last updated on 24/May/20

$$\mathrm{let}2\mathrm{p},2\mathrm{q},2\mathrm{r}\mathrm{\&}2\mathrm{s}\mathrm{are}\mathrm{the}\mathrm{complex}\mathrm{numbers}$$$$\mathrm{represents}\mathrm{the}\mathrm{four}\mathrm{sides}\mathrm{of}\mathrm{the}\mathrm{quadrilateral}$$$$\therefore \mathrm{p}+\mathrm{q}+\mathrm{r}+\mathrm{s}=0$$$$\mathrm{now}\mathrm{A}=\mathrm{p}+\mathbf{i}\mathrm{p}=(1+\mathbf{i})\mathrm{p}$$$$\mathrm{B}=2\mathrm{p}+(1+\mathbf{i})\mathrm{q}\mathrm{C}=2\mathrm{p}+2\mathrm{q}+(1+\mathbf{i})\mathrm{r}$$$$\mathrm{D}=2\mathrm{p}+2\mathrm{q}+2\mathrm{r}+(1+\mathbf{i})\mathrm{s}$$$$\mathrm{AC}=\mathrm{p}+2\mathrm{q}+\mathrm{r}+\mathbf{i}(\mathrm{r}-\mathrm{p})$$$$\mathrm{BD}=\mathrm{D}-\mathrm{B}=\mathrm{q}+2\mathrm{r}+\mathrm{s}+\mathbf{i}(\mathrm{s}-\mathrm{q})$$$$\mathrm{BD}+\mathbf{i}\mathrm{AC}=(\mathrm{p}+\mathrm{q}+\mathrm{r}+\mathrm{s})+\mathbf{i}(\mathrm{p}+\mathrm{q}+\mathrm{r}+\mathrm{s})=0$$$$\Rightarrow \mathbf{i}\mathrm{BD}=\mathrm{AC}\Rightarrow \mathrm{BD}=\mathrm{AC}\mathrm{\&}\mathrm{AC}\mathrm{\perp }\mathrm{BD}$$

Commented by PRITHWISH SEN 2 last updated on 24/May/20

Commented by mr W last updated on 24/May/20

$$thankssir!$$

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