ABCD is a quadrilateral. E, F, G, and H are midpoints of sides AB, BC, CD, and DA respectively. DEF and BGH are equilateral triangles.
Find the angles inside the quadrilateral ABCD.
Interesting question. Not too difficult, if you try it.
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see diagram.
EF = AC/2 = GH = 2x.
So FD = 2x
FG = BD/2 = EH = 2 y.
Triangle IFD, FD² = x² + 9 y²
so 4x² = x² + 9 y²
x/y = √3
In triangle BIF or BIE: angle B = 2* tan⁻¹ (x/y) = 2 *60⁰ = 120⁰
Angle D = angle B = 120⁰. Angle A = angle C = 60⁰.
We can observe that ABCD is a Rhombus. EFGH is a rectangle.
EF = AC/2 = GH = 2x.
So FD = 2x
FG = BD/2 = EH = 2 y.
Triangle IFD, FD² = x² + 9 y²
so 4x² = x² + 9 y²
x/y = √3
In triangle BIF or BIE: angle B = 2* tan⁻¹ (x/y) = 2 *60⁰ = 120⁰
Angle D = angle B = 120⁰. Angle A = angle C = 60⁰.
We can observe that ABCD is a Rhombus. EFGH is a rectangle.
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kvnmurty:
click on red heart thanks above pls
Great explanation sir
Awesome Answer!
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