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Question Number 181125 by depressiveshrek last updated on 21/Nov/22

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

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

Commented by mr W last updated on 22/Nov/22

please check the question!  intersection point of heights of  ΔADE = orthocenter of ΔADE ?  it seems that orthocenter of ΔADE  is not the midpoint of BC. you can  check when you make a drawing.

$${please}\:{check}\:{the}\:{question}! \\ $$$${intersection}\:{point}\:{of}\:{heights}\:{of} \\ $$$$\Delta{ADE}\:=\:{orthocenter}\:{of}\:\Delta{ADE}\:? \\ $$$${it}\:{seems}\:{that}\:{orthocenter}\:{of}\:\Delta{ADE} \\ $$$${is}\:{not}\:{the}\:{midpoint}\:{of}\:{BC}.\:{you}\:{can} \\ $$$${check}\:{when}\:{you}\:{make}\:{a}\:{drawing}. \\ $$

Commented by depressiveshrek last updated on 22/Nov/22

I also find the question to be confusing,  the intersection is clearly not the  midpoint of BC

$${I}\:{also}\:{find}\:{the}\:{question}\:{to}\:{be}\:{confusing}, \\ $$$${the}\:{intersection}\:{is}\:{clearly}\:{not}\:{the} \\ $$$${midpoint}\:{of}\:{BC} \\ $$

Commented by mr W last updated on 22/Nov/22

i checked again. the question is  correct. sorry!

$${i}\:{checked}\:{again}.\:{the}\:{question}\:{is} \\ $$$${correct}.\:{sorry}! \\ $$

Answered by mr W last updated on 22/Nov/22

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