Right is right
Two right isosceles triangles are inscribed in a quadrilateral. Their right angles are at opposing vertices of the quadrilateral. Their other vertices are midpoint of its sides. Find the largest angle of the quadrilateral.
Details: A F = F B , B G = G C , C H = H D , D E = E A
∠ E C F = 9 0 ∘ , ∠ H A G = 9 0 ∘ , A H = A G , C E = C F
Drawing is not to scale.
Inspiration: Just one step to regularity of quadrilateral by Rohit Camfar.
The answer is 143.13.
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2 solutions
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Little CONSTRUCTION.
From the statement of the problem, it is not immediately known that DB=GH. That is part of what needs to be shown to be true.
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Actually, DB is not equal to GH. It is twice. I have used fact that DB || GH. That is obvious through Midpoint theorem.The only thing which I have escaped is to prove that ABCD is a rhombus, which is very obvious in this situatiob and can be proved through congruency.
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Thanks for straightening me out. Now I can see it.
I meant to say parallel, sorry. DB parallel to GH.
Its unnecessarily confusing to make the drawing intentionally incorrect.
We can see that
A
B
=
B
C
=
C
D
. The first equality comes from the fact that
E
and
F
are midpoints of
A
J
and
A
K
, the second from symmetry between triangles
A
H
G
and
L
E
F
.
The entire length of A D = D G = B F . The first equality is from the position of the two segments in the right isosceles triangle A H G , the other from the same symmetry as before.
So in the triangle A B F the angle B A F = a r c t a n ( x 3 x ) = a r c t a n 3 = 7 1 . 5 6 5 ∘ , giving us the angle J A K = 2 × 7 1 . 5 6 5 ∘ = 1 4 3 . 1 3 ∘ . .