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Question Number 23253 by ajfour last updated on 28/Oct/17

Commented by ajfour last updated on 28/Oct/17

$$Q.23251\left(solution\right)$$

Commented by math solver last updated on 28/Oct/17

$$\mathrm{i}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{get}\mathrm{what}\mathrm{you}\mathrm{write}\mathrm{in}\mathrm{diagram}$$

Commented by math solver last updated on 28/Oct/17

$$\mathrm{how}\mathrm{did}\mathrm{you}\mathrm{write}\mathrm{ka}/\mathrm{a}+4,4\mathrm{a}/\mathrm{a}+4$$$$\mathrm{and}\mathrm{all}\mathrm{such}\mathrm{lengths}.$$

Commented by math solver last updated on 28/Oct/17

$$??$$

Commented by ajfour last updated on 28/Oct/17

$$theyare\frac{ka}{k+4}and\frac{4a}{k+4},etc.$$

Commented by math solver last updated on 28/Oct/17

$$\mathrm{yes}\mathrm{i}\mathrm{am}\mathrm{asking}\mathrm{these}\mathrm{lengths}\mathrm{only}$$$$\mathrm{it}\mathrm{is}\mathrm{seem}\mathrm{to}\mathrm{written}\mathrm{similar}\mathrm{to}$$$$\mathrm{that}\mathrm{in}\mathrm{case}\mathrm{of}\mathrm{incentre}\mathrm{also}.\mathrm{plz}$$$$\mathrm{explain}\mathrm{it}?$$

Commented by ajfour last updated on 28/Oct/17

$$trytofocusonQ.23262,itis$$$$foryou.$$

Commented by math solver last updated on 28/Oct/17

$$\mathrm{thank}\mathrm{you}\mathrm{sir}!$$

Answered by ajfour last updated on 28/Oct/17

$${BC}^2={AB}^2+{AC}^2$$$$\Rightarrow {(k+4)}^2{r}^2=a^2+b^2....\left(i\right)$$$${AP}^2={\left(\frac{ka}{k+4}\right)}^2+{\left(\frac{4b}{k+4}\right)}^2={\left(3r\right)}^2$$$$\Rightarrow k^2{a}^2+16b^2=9{(k+4)}^2{r}^2.....\left(ii\right)$$$$using\left(i\right)in\left(ii\right):$$$$k^2{a}^2+16b^2=9(a^2+b^2)$$$$\Rightarrow k=\sqrt{\frac{9a^2-7b^2}{a^2}}$$$$=\sqrt{9-7{\left(\frac{b}{a}\right)}^2}$$$$asa=\sqrt{7}andb=\sqrt{5}weget$$$$\mathit{k}=2$$$$\Rightarrow \frac{\mathit{B}\mathit{P}}{\mathit{P}\mathit{C}}=\frac{4\mathit{r}}{\mathit{k}\mathit{r}}=\frac{4}{\mathit{k}}=\frac{4}{2}=\frac{2}{1}.$$$$$$

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