Question Number 15987 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Jun/17 | ||
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Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Jun/17 | ||
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$${H},{is}\:{a}\:{point}\:{on}\:{circle}\:{with}\:{centerpoint}:{A}. \\ $$$$\:\:{prove}\:\:{that}: \\ $$$$\:\:\:\:\:\:\:\:\angle\:{DHF}=\mathrm{135}^{\bullet} \:. \\ $$ | ||
Answered by ajfour last updated on 16/Jun/17 | ||
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$$\angle{DHF}=\frac{\mathrm{1}}{\mathrm{2}}\left({reflex}\angle{DAF}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{270}°\right)\:=\:\mathrm{135}°\:\:. \\ $$ | ||