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Question Number 29541 by naka3546 last updated on 09/Feb/18

Commented by naka3546 last updated on 09/Feb/18

Prove  that  DF  is  perpendicular  to  IC .

$${Prove}\:\:{that}\:\:{DF}\:\:{is}\:\:{perpendicular}\:\:{to}\:\:{IC}\:. \\ $$

Commented by ajfour last updated on 09/Feb/18

Commented by ajfour last updated on 09/Feb/18

tan θ=((c−acos α−asin α)/(acos α−asin α))  tan φ=((bcos β−bsin β)/(c−bcos β−bsin β))  and since   acos α+bcos β=c  also                asin α=bsin β  therefore    tan 𝛉 = tan 𝛗        so   DF ⊥ to IC  .

$$\mathrm{tan}\:\theta=\frac{{c}−{a}\mathrm{cos}\:\alpha−{a}\mathrm{sin}\:\alpha}{{a}\mathrm{cos}\:\alpha−{a}\mathrm{sin}\:\alpha} \\ $$$$\mathrm{tan}\:\phi=\frac{{b}\mathrm{cos}\:\beta−{b}\mathrm{sin}\:\beta}{{c}−{b}\mathrm{cos}\:\beta−{b}\mathrm{sin}\:\beta} \\ $$$${and}\:{since}\:\:\:{a}\mathrm{cos}\:\alpha+{b}\mathrm{cos}\:\beta={c} \\ $$$${also}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}\mathrm{sin}\:\alpha={b}\mathrm{sin}\:\beta \\ $$$${therefore}\:\:\:\:\mathrm{tan}\:\boldsymbol{\theta}\:=\:\mathrm{tan}\:\boldsymbol{\phi} \\ $$$$\:\:\:\:\:\:{so}\:\:\:{DF}\:\bot\:{to}\:{IC}\:\:. \\ $$

Commented by mrW2 last updated on 09/Feb/18

Great!

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