【Haichuan Teahouse】Brain-teaser Series: Neither a trapezoid nor a parallelogram
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In quadrilateral ABCD, BA=2DA=6, BC=CD, ∠DCB=120 degrees; draw AC, and find the area of triangle ABC.
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This problem lacks necessary conditions; I was able to come up with at least 3 shapes, each with a different area.
There must be other data as well; otherwise, countless graphs could be created.
I also feel it’s not quite right. Could you send over the image you created to take a look? But it seems like there’s only one option that meets the requirements?
The last edit to this post was made by Thank goodness you're here on 2021-7-26 at 17:00. No screenshots will be used; I’ll explain my approach using CAD: 1. BC=CD; draw a circle with center C and any radius, then create an angle BCD of 120 degrees, and keep it aside for later use； 2. With A as the center, draw a circle with a radius of 3 ； 3. On the circle with a radius of 3, arbitrarily select the position of point D, and connect BD (AB=6; point B can be determined using point A) ； 4. Since BC = CD, the center C must lie on the perpendicular bisector of BD; therefore, draw the perpendicular bisector of BD ； 5. By using the AL adjustment command, a parallel line to BC can be obtained, which can then be moved to point B ； 6. The intersection of the perpendicular bisectors of BC and BD is the center C of the circle. By arbitrarily changing the position of point D and using the AL adjustment command, multiple positions for point C can be obtained, along with the area values of the corresponding triangles ABC.
This old guy doesn’t know where he got those questions; P