Let ABCD be an isosceles trapezoid, with bases AB and CD. and A circle is inscribed in the trapezoid, as shown below. (In other words, the circle is tangent to all the sides of the trapezoid.) The length of base AB is 2x and the length of base CD is 2y. Prove that the radius of the inscribed circle is sqrt(xy)
https://gyazo.com/4dd0a1236ad5c782bc9d82015f5cba3b
Not sure how to upload image, just a trapezoid with top base AB and lower base DC respectively, with a circle inscribed in it so the circle is tangent to every side.
+1768 Let O be the center of the inscribed circle. Let E be the midpoint of AB, and let F be the midpoint of CD. Draw segments AO, BO, CO, and DO. Since ABCD is an isosceles trapezoid, ∠ABE=∠CBE and ∠DCE=∠EDF, so triangles ABE, CBE, DCE, and EDF are all isosceles.
Since AE=BE, CE=DE, and AO=BO, triangles AEO, BEO, CEO, and DEO are all congruent. Therefore, OA=OB=OC=OD=r, where r is the radius of the inscribed circle.
Let E′ be the foot of the altitude from E to AB, and let F′ be the foot of the altitude from F to CD. Since triangles AEE′ and BEE′ are both right isosceles triangles, AE′=BE′=r, and similarly, CF′=DF′=r. Therefore, the sum of the lengths of the segments from E to D is: [ EE' + ED + DF' = 2r + x + 2r = x + 4r. ]
Similarly, the sum of the lengths of the segments from F to B is y+4r. Since ABCD is an isosceles trapezoid, the sum of the lengths of the segments from E to D is equal to the sum of the lengths of the segments from F to B. Therefore, x+4r=y+4r, so x=y. Hence, the radius of the inscribed circle is sqrt(xy).
