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Question Number 163656 by mnjuly1970 last updated on 09/Jan/22

Answered by mr W last updated on 09/Jan/22

Commented by mr W last updated on 09/Jan/22

$$m=\sqrt{a^2-h_1^2}+\sqrt{b^2-h_1^2}$$$$m-\sqrt{a^2-h_1^2}=\sqrt{b^2-h_1^2}$$$$m^2+a^2-h_1^2-2m\sqrt{a^2-h_1^2}=b^2-h_1^2$$$$m^2+a^2-b^2=2m\sqrt{a^2-h_1^2}$$$$\Rightarrow \sqrt{a^2-h_1^2}=\frac{m^2+a^2-b^2}{2m}$$$$similarly$$$$\Rightarrow \sqrt{p^2-h_2^2}=\frac{n^2+p^2-q^2}{2n}$$$$$$$$\sqrt{p^2-h_2^2}-\sqrt{a^2-h_1^2}=\frac{n-m}{2}$$$$\frac{n^2+p^2-q^2}{2n}-\frac{m^2+a^2-b^2}{2m}=\frac{n-m}{2}$$$$\frac{p^2-q^2}{n}-\frac{a^2-b^2}{m}=0$$$$\Rightarrow \frac{a^2-b^2}{p^2-q^2}=\frac{m}{n}$$

Commented by mnjuly1970 last updated on 09/Jan/22

$$verynicesolution\mathrm{sir}\mathrm{W}...\mathrm{thank}$$$$\mathrm{you}\mathrm{so}\mathrm{much}...$$

Commented by Tawa11 last updated on 09/Jan/22

$$\mathrm{Great}\mathrm{sir}.$$

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