algebra-precalculusgeometry
Let $ABCD$ be a trapezium with parallel sides $AB$ and $CD$ such that the circle $S$ with $AB$ as its diameter touches $CD$. Further, the circle $S$ passes through the midpoints of the diagonals $AC$ and $BD$ of the trapezium. The smallest angle of the trapezium is ______.
Can anyone please tell me how to proceed?
I think some more information needed to solve the problem.
The midline divides the trapezoid into two equally high trapezoids. Its length is also the mean of the length of the two parallel sides. (You can see that by extending the nonparallel sides until they intersect in a point. The two parallel sides and the midline then forms the base line in three similar triangles).
Call the long parallel side $AB$, and the short one $CD$ with $AD$ and $BC$ being the two other sides of the trapezoid. Further let $r$ be the radius of the circle. The trapezoid then has height $2r$. Further, let $E, F, G, H$ be the places the circle touches the trapezoid at $AB$, $BC$, $CD$ and $AD$ respectively.
The hypothesis about the midline dividing the area (note that the two parts are also trapezoids) then dictates that $$ 3\frac{AB + \frac{AB + CD}{2}}{2}r = 5\frac{CD + \frac{AB + CD}{2}}{2}r \\ {3}AB + 3\frac{AB + CD}{2} -5CD -5\frac{AB+CD}{2} = 0\\ 2AB - 6CD = 0\\ AB = 3CD $$ so the midline has length $2CD$.
Since the sides touch the circle we must have the following: $$ AE = AH\\ BE = BF\\ CF = CG\\ DG = DH $$ Since $DH + AH + BF + CF = 16$, then we must have $AE+BE + CG + DG = 16$. But this also equals $4CD$, so $CD = 4$. The long parallel side $AB$ must therefore be $12$.
1) Let the line $MO$ intersects the sides of the trapezoid at the points $E$ and $F$. Then $$\triangle BME \sim \triangle AMF \Rightarrow\frac {BE}{AF}=\frac {ME}{MF}$$
2) $$\triangle EMC \sim \triangle FMD \Rightarrow\frac {ME}{MF}=\frac {EC}{FD}\Rightarrow\frac {BE}{AF}=\frac {EC}{FD}$$
3) $$\triangle AOF \sim \triangle COE \Rightarrow\frac {EC}{AF}=\frac {EO}{OF}$$
$$\triangle BEO \sim \triangle DFO \Rightarrow\frac {BE}{FD}=\frac {EO}{OF} \Rightarrow\frac {BE}{FD}=\frac {EC}{AF}$$ So $$\frac {BE}{AF}=\frac {EC}{FD}$$ and $$\frac{BE}{FD}=\frac {EC}{AF}$$
Then the four points $M,E,O,F$ lie on a straight line
Best Answer
The triangles AFO and ACB are similar since AO = OB, AF = FC and shared angle $\alpha$, , which leads to BC = 2OF = 2r. From the right triangle BEC, we have $\sin\angle$BCE = BE/BC = $\frac12$. Thus, the angle is 30$^\circ$.