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?

Guest Nov 26, 2019

#2**+1 **

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

Melody Nov 27, 2019

#3**0 **

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...**

Melody Nov 27, 2019

#5**+2 **

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).

Tiggsy Nov 28, 2019

#9**+2 **

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,

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

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

.Tiggsy Nov 29, 2019