MHB Calculating the Length of BC: A Geometric Exploration

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The discussion centers on determining the length of segment BC in a geometric problem involving point E, which lies on the angle bisector of angle A and is not on BC. It is established that the length of BC is not unique and can vary based on the diagram drawn by the student, with examples suggesting BC could be greater than 16. The conversation emphasizes the importance of student analysis in open-ended problems. A specific case is presented where BC is given as 50/3, leading to a calculation of the sum of the triangle's side lengths. The conclusion confirms that the calculated side lengths are correct based on the provided values.
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Albert said:
Find length of BC:

View attachment 708
Hello Albert,

In the above problem can you please clarify if E lies on BC or not? If not, then how is E defined other than than AE=10.
 
Point E is on angle bisector of angle A, and AE=10

E is not on BC (from the diagram given)
 
Albert said:
Point E is on angle bisector of angle A, and AE=10

E is not on BC (from the diagram given)
But then I think BC doesn't have a unique length. Take BC =16 and the mid point of BC be D. Point E is on the side of BC different from that of A and ED=2 with ED perpendicular to BC. In fact any value of BC>16 can provide a configuration confirming to the hypothesis if the question.
 
yes, the length of BC is not unique ,it depends on how student draw the diagram

sometimes we plot an open problem ,and want to test the abilities of students in analyzing

now using the diagram given if I said BC=$\dfrac {50}{3}$

what wiil be the sum of the three side lengths of triangle ABC ?
 
Last edited:
Albert said:
yes, the length of BC is not unique ,it depends on how student draw the diagram

sometimes we plot an open problem ,and want to test the abilities of students in analyzing

now using the diagram given if I said BC=$\dfrac {50}{3}$

what wiil be the sum of the three side lengths of triangle ABC ?
$10$ and $40/3$. Is this correct?
 
yes,you got it :)
 
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