help

Is there a diagram?

A semicircle is inscribed in a quadrilateral ABCD in such a way that the midpoint of BC coincides with the center of the semicircle, and the diameter of the semicircle lies along a portion of BC. If AB = 4 and CD = 5, what is BC?

I have not solved this problem.

I have only drawn an initial diagram.

Here it is.  W is the centre of the circle and also the centre of side BC of the quadrilateral

Maybe you need to use some obscure circle geomtery ratio  idk

It is driving me nuts. Can someone please show me how do do it...

I get an answer of 4sqrt(5).

I need to check my algebra and, if correct, I'll try to a post a solution tomorrow, (no time at the moment).

See diagram from previous post.

Triangles BEO and CGO are congruent.

Let BE = x , the angle at E is a right angle so  x = (R + h)cos(theta).

EA = 5 - x, and if AD is to be a tangent to the semi circle, then the line from A to the point of tangency will have length 5 - x also.

Similarly on the RHS, the line from D to the point of tangency will have length 4 - x.

So, if AD is to be a tangent to the semi circle, it will have a length 9 - 2x = 9 - 2(R + h)cos(theta).

Now move into co-ordinate geometry mode.

Take BC to be the horizontal axis, with O as the origin.

Point A will have an x co-ordinate   -(R + h) + 5cos(theta) and a y co-ordinate 5sin(theta).

Point D will have an x co-ordinate    (R + h) - 4cos(theta) and a y co-ordinate 4sin(theta).

So,

\(\displaystyle AD^{2}= \{9-2(R+h)\cos\theta\}^{2}=\{(R+h)-4\cos\theta-\{-(R+h)+5\cos\theta\}\}^{2}+(4\sin\theta-5\sin\theta)^{2},\\ 81-36(R+h)\cos\theta+4(R+h)^{2}\cos^{2}\theta=4(R+h)^{2}-36(R+h)\cos\theta+81\cos^{2}\theta+\sin^{2}\theta,\\ 81+4(R+h)^{2}\cos^{2}\theta-4(R+h)^{2}-81\cos^{2}\theta-\sin^{2}\theta = 0, \\ 4(R+h)^{2}(\cos^{2}\theta-1)+81\sin^{2}\theta -\sin^{2}\theta=0,\\ 4(R+h)^{2}(-\sin^{2}\theta)+80\sin^{2}\theta=0,\\ (R+h)^{2}=20,\\ R+h =\sqrt{20}=2\sqrt{5} . \)

\(\displaystyle \text{BC}=2(R+h)=4\sqrt{5}.\)

Thanks very much Tiggsy.  I have not studied it yet but I will.