incircle trig problem
Hi, would any one be able to help me solve the following problem?
ABC is a triangle in which none of the angles is obtuse. The perpendicular AD from A to BC is produced to meet the circumcircle of the triangle at E. If D is equidistant from A and E prove that the triangle must be right-angled. If, alternatively, the incentre of the triangle is equidistant from A and E, prove that cos B + cos C = 1.
I can see that D must lie on the diameter as must B and C, so A must be 90 degrees. The second part of the question is where I am stumped. I have been trying to find an expression that constrains the centre of the incircle to be equidistant from A and E and transform that expression to the required one. However I don't know where to begin.
ABC is a triangle in which none of the angles is obtuse. The perpendicular AD from A to BC is produced to meet the circumcircle of the triangle at E. If D is equidistant from A and E prove that the triangle must be right-angled. If, alternatively, the incentre of the triangle is equidistant from A and E, prove that cos B + cos C = 1.
I can see that D must lie on the diameter as must B and C, so A must be 90 degrees. The second part of the question is where I am stumped. I have been trying to find an expression that constrains the centre of the incircle to be equidistant from A and E and transform that expression to the required one. However I don't know where to begin.
Original post by Stanley appleton
Hi, would any one be able to help me solve the following problem?
ABC is a triangle in which none of the angles is obtuse. The perpendicular AD from A to BC is produced to meet the circumcircle of the triangle at E. If D is equidistant from A and E prove that the triangle must be right-angled. If, alternatively, the incentre of the triangle is equidistant from A and E, prove that cos B + cos C = 1.
I can see that D must lie on the diameter as must B and C, so A must be 90 degrees. The second part of the question is where I am stumped. I have been trying to find an expression that constrains the centre of the incircle to be equidistant from A and E and transform that expression to the required one. However I don't know where to begin.
ABC is a triangle in which none of the angles is obtuse. The perpendicular AD from A to BC is produced to meet the circumcircle of the triangle at E. If D is equidistant from A and E prove that the triangle must be right-angled. If, alternatively, the incentre of the triangle is equidistant from A and E, prove that cos B + cos C = 1.
I can see that D must lie on the diameter as must B and C, so A must be 90 degrees. The second part of the question is where I am stumped. I have been trying to find an expression that constrains the centre of the incircle to be equidistant from A and E and transform that expression to the required one. However I don't know where to begin.
I hope some proper mainland Greek or Australian is around because there are very few geometers around in the UK.
I have even forgotten what incentre, circumcentre, orthocentre etc mean.
[I will try but not tonight as brain has seized...]
(edited 9 years ago)
Original post by Stanley appleton
Hi, would any one be able to help me solve the following problem?
ABC is a triangle in which none of the angles is obtuse. The perpendicular AD from A to BC is produced to meet the circumcircle of the triangle at E. If D is equidistant from A and E prove that the triangle must be right-angled. If, alternatively, the incentre of the triangle is equidistant from A and E, prove that cos B + cos C = 1.
I can see that D must lie on the diameter as must B and C, so A must be 90 degrees. The second part of the question is where I am stumped. I have been trying to find an expression that constrains the centre of the incircle to be equidistant from A and E and transform that expression to the required one. However I don't know where to begin.
ABC is a triangle in which none of the angles is obtuse. The perpendicular AD from A to BC is produced to meet the circumcircle of the triangle at E. If D is equidistant from A and E prove that the triangle must be right-angled. If, alternatively, the incentre of the triangle is equidistant from A and E, prove that cos B + cos C = 1.
I can see that D must lie on the diameter as must B and C, so A must be 90 degrees. The second part of the question is where I am stumped. I have been trying to find an expression that constrains the centre of the incircle to be equidistant from A and E and transform that expression to the required one. However I don't know where to begin.
You can try by assuming coordinates, and finding a relation between x and y, but the process is very long-winding. SOT was always my weak point, I'll go through my notes again and see if something can be done.
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