Challenging Geometry Puzzle: Problem 1591. Share your solution by posting it in the comment box provided.
Audience: Mathematics Education - K-12 Schools, Honors Geometry, and College Level.
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Draw perpendicular CE from C to AD, E on AD
ReplyDeleteDraw perpendicular BF from B to AC, F on AC
Right Triangles ABF, CBF & CED are all congruent (AB = BC = CD, 90, 5@) and
So AC = 2.CE
It follows that ACE is a 30-60-90 Triangle
So < BAC = 7@ - 30
Hence (7@ - 30) + 5@ = 90 from Triangle BAD
Therefore @ = 10 and < D = 50
Sumith Peiris
Moratuwa
Sri Lanka
https://photos.app.goo.gl/JB2RXPwcvKovVaHz8
ReplyDeleteDenote AB=BC=CD=r and a= alpha
Triangle ABC is isosceles=> ∡(BAC)= 90-5a
And ∡(CAD)= 12a-9 => AC= 2.r.sin(5a)
In triangle ACD r/sin(12a-90)= AC/sin(5a)= 2r
So sin(12a-90)= ½ => cos(12a)= -½ => a= 10 degrees
So Angle D= 50
Place point E on side AD so that BA=BE.
ReplyDeleteLet angle BEC=angle BCE=θ
The sum of the interior angles of quadrilateral ABCE is 360°, so
10α+7α+7α+2θ=360°
∴θ=180°-12α
The sum of the interior angles of quadrilateral ABCD is 360°, so
5α+7α+10α+(180°-12α)+angle CDE=360°
Angle CDE=180°-10α
∴Angle CED=5α
From the above, triangle BCE is an equilateral triangle, so
180°-12α=60°
α=10°
Angle D=5α=50°
Hi Antonio hope you received my Solution to Problem 1591 sent a few days ago
ReplyDeleteIn case you did not receive it, here it is again
Drop perpendiculars BE to AC and CF to AD
Triangles ABE, CBE and CFD are congruent ASA and hence
AC = 2.CF
It follows that ACF is a 30-60-90 Triangle
So in Triangle ABE,
(7@ - 30) + 5@ = 90
Therefore @ = 10 and
<D = 50
Sumith Peiris
Moratuwa
Sri Lanka
Take point E on line segment AD where angle BAE = angle BEA
ReplyDeleteTriangle EBA is an isosceles triangle with BE = BC
Let angle BEC = angle BCE = x
The sum of the interior angles of quadrilateral ABCE is 360°, so
10α + 7α + 7α + 2x = 360°
Therefore, x = 180° - 12α
Angle BEA + angle BEC + angle CED = 180°
7α + (180° - 12α) + angle CED = 180°
so angle CED = 5α
Therefore, triangle BEC is an equilateral triangle
Solving 180° - 12α = 60° gives
α = 10°
Angle D = 5α = 50°