AB = 6, AD = 3, DE = 2, and BDE = 90 degrees, How to construct this triangle?

Question

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.

Update:

The triangle is unique with the given inputs.

Update 2:

ACB = 90 degrees

Update 3:

AB≠CD

dawonzcher · Answer

I'm also confused at this. (Anyway, take a chance to have a look in the way i see it. It might help.)
If you can notice, segment DE is a radius of another smaller circle whose center is at point D, and CD is also same radius. Hence DE = 2 = CD, then it makes AC = AD + DC = 5. Then it become simple to construct the inner triangle.

Now, you can solve for BC using Pythagorean theorem:
(AB)^2 = (AC)^2 + (BC)^2
(BC)^2 = (AB)^2 - (AC)^2
BC= (6^2 - 5^2)^1/2
BC = (36 - 25)^1/2
BC = sqrt(11) = 3.317  <===Answer

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AB = 6, AD = 3, DE = 2, and BDE = 90 degrees, How to construct this triangle?

1 Answer