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AB = 6, AD = 3, DE = 2, and BDE = 90 degrees, ACB = 90 degrees. How to construct this triangle?
This question was put up by someone few days back asking for measure of BC. My problem is that I could not even construct the triangle. The locus of E is from the tangent point from A to the circle with centre D & the intersecting point of the circle and AC. (Yellow arc) In this diagram BDE is not 90.
NOTE:-- AB≠CD
The triangle is unique with the given inputs.
Sorry, it is DE ≠ CD.
1 Answer
- ?Lv 66 years agoFavorite Answer
On x-y plane, put A(0,0) and B(6,0).
Since AD=3, the coordinates of D is (x,sqrt(9-x^2)).
Draw a perpendicular line from D toward AB,
the point of intersction is named F.
Since angle EDB is a right angle,
triangle EDF and DBF are similar right triangles.
So EF:DF=DF:BF,
and let EF=u, DF=sqrt(9-x^2), BF=6-x,
then u:sqrt(9-x^2)=sqrt(9-x^2):(6-x),
u(6-x)=9-x^2
u=(9-x^2)/(6-x) ---(1)
At triangle EDF EF^2+DF^2=DE^2=4,
then u^2+(9-x^2)=4.
And from (1),
(9-x^2)^2/(6-x)^2+(9-x^2)=4
(9-x^2)^2/(6-x)^2-x^2+5=0
Multiply this equation by (6-x)^2,
12x^3-49x^2-60x+261=0
x=2.41 (applox.)
Thus D’s coordinates are found.
After this, to find C's coordinates is easy.