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GeometryQuestion and Answers: Page 45

Question Number 181151    Answers: 4   Comments: 0

Question Number 181128    Answers: 0   Comments: 2

Question Number 181125    Answers: 1   Comments: 3

Let the acute triangle ΔABC have an outer circumscribed circle, whose tangents at the points B and C intersect at point P. Let D and E be the projections of perpendicular lines from point P on AC and AB. Prove that the interdection point of the heights of ΔADE is the midpoint of BC

$${Let}\:{the}\:{acute}\:{triangle}\:\Delta{ABC}\:\:{have} \\ $$$${an}\:{outer}\:{circumscribed}\:{circle}, \\ $$$${whose}\:{tangents}\:{at}\:{the}\:{points}\:{B}\:{and}\:{C} \\ $$$${intersect}\:{at}\:{point}\:{P}.\:{Let}\:{D}\:{and}\:{E}\:{be} \\ $$$${the}\:{projections}\:{of}\:{perpendicular} \\ $$$${lines}\:{from}\:{point}\:{P}\:{on}\:{AC}\:{and}\:{AB}. \\ $$$${Prove}\:{that}\:{the}\:{interdection}\:{point}\:{of} \\ $$$${the}\:{heights}\:{of}\:\Delta{ADE}\:{is}\:{the}\:{midpoint} \\ $$$${of}\:{BC} \\ $$

Question Number 181085    Answers: 1   Comments: 3

Question Number 181070    Answers: 2   Comments: 2

Question Number 181055    Answers: 2   Comments: 14

Question Number 181051    Answers: 1   Comments: 0

Question Number 181019    Answers: 1   Comments: 0

Question Number 180998    Answers: 2   Comments: 0

Question Number 180939    Answers: 0   Comments: 2

Question Number 180931    Answers: 0   Comments: 2

Question Number 180927    Answers: 2   Comments: 1

Question Number 180882    Answers: 1   Comments: 1

Question Number 180897    Answers: 1   Comments: 5

Question Number 180813    Answers: 1   Comments: 5

Question Number 180756    Answers: 3   Comments: 1

Question Number 180623    Answers: 1   Comments: 0

Question Number 180615    Answers: 3   Comments: 0

Question Number 180582    Answers: 1   Comments: 0

Question Number 180547    Answers: 1   Comments: 0

Question Number 180510    Answers: 2   Comments: 1

Question Number 180459    Answers: 2   Comments: 0

Question Number 180458    Answers: 2   Comments: 0

Question Number 180423    Answers: 2   Comments: 4

Question Number 180400    Answers: 2   Comments: 1

Area BEF=AreaCDF Prouve AD×BE=AE×CD

$$\mathrm{Area}\:\mathrm{BEF}=\mathrm{AreaCDF} \\ $$$$\mathrm{Prouve}\:\:\mathrm{AD}×\mathrm{BE}=\mathrm{AE}×\mathrm{CD} \\ $$

Question Number 180365    Answers: 1   Comments: 0

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