this is supposed to be a non-calculator question?

Duke has drawn my attention to this wonderful problem. His solution is mind-boggling and I always wonder how he can think so deeply. I was inspired by his solution and have an alternate method.

Refer to the figure

http://www.flickr.com/photos/52771834@N/43154893…

ABCD is the given quadrilateral.

CN is bisector of ∠ BCD.

A’BCD is the image of ABCD in CN.

It can easily be proved that

∠ BCA = 55°, ∠ ACN = 30°

Δs ACN and A’CN are congruent right triangles.

=> Δ AA’N is equilateral.

It can be proved that AD and A’D are angle bisectors of Δ AA’C

=> ∠ CAD = 30°

=> ∠ BAD

= ∠ BAC + ∠CAD

= 55° + 30°

= 85°.

Edit:

My son, Kuntal, who is a networking professional with Cisco Systems and despite being out of touch with the subject, has found a simpler logic based solution which I shall post after the figure is drawn. Please keep the question open till then.

Kuntal’s solution:

Refer to the figure:

http://www.flickr.com/photos/52771834@N/43173048…

[All the angles shown in the figure are in degrees.]

ABCD is the given quadrilateral.

Draw equilateral ΔBEC.

E may lie inside or outside quadrilateral ABCD or on side AD. We assume that it is inside ABCD and prove that this is not possible.

With the given quadrilateral and the construction, different angles are as shown in the figure.

=> Sum of all angles of the quadrilateral

= α° + 85° + 70° + 170° + 35° + β°

= 360° + α° + β°

=> α° + β° = 0.

This can be proved even if E were outside ABCD.

=> E is on AD so that α° + β° = 0.

=> ∠ BAD = 85°.

2

My TU (already given) for Duke sir, Madhukar sir

and Rozeta for their excellent answers.

I will be glad to see myself among geniuses. Here is the another way to this nice sum.

Please have a look to the link given below.

http://i854.photobucket.com/albums/ab104/RakeshDub…

Given that : AB = BC = CD and ∠ ABC = 70° ; ∠ BCD = 170°.

► Constructing CE // BA (point E is on the circumference of the circle with center C)

Now, ∠ BCE = ∠ BAE = 110° – – {adjacent angles of a rhombus are supplementary }

► ∠ DCE = 60° and hence Δ DCE will be an equlateral triangle.

►∠ DEA = 130° – – – { 60° + 70° }

►∠ EDA = ∠EAD = 25° – – { Δ EDA is an isosceles Δ having sides ED = EA }

Therefore, ∠ BAD = ∠EAB – ∠EAD ==> 110° – 25°

∠ BAD = 85° ► ( Answer )

==========================================

Edit : It can be done without using circle (which, I have drawn is only for the sake of simplicity). For that, From point C,draw a line parallel to BA and from point D, draw an arc of length DC, which meets the // line drawn at point E. Join EA. DCE will be an equilateral triangle and ABCE will be a rhombus.

I suppose ABCD is convex? No calculator is necessary indeed – the answer is exactly 85° – follow the link below to see a picture and an explanation:

http://farm5.static.flickr.com/4022/4313222727_3da…

Let the diagonals AC and BD intersect in E. ABC and BCD are isosceles, hence angles CAB and ACB are 55°, angles DBC and BDC are 5°. Draw a line BA’ such that

angle(ABA’) = angle(CBE) = 5° (A’ on AC, disregard B’ on the picture)

Draw a line CD’ (D’ on BD) such that angle(DCD’) = 55°

Triangles ABA’, CBE and CDD’ are congruent (green side and 5° and 55° angles), hence segments A’B, BE and DD’ (red) are equal. Triangle A’BE is isosceles with 60° angle, hence equilateral, or

|A’E| = |A’B| = |BE| = |DD’ |

Triangle CD’E (cyan) is also equilateral (2 blue sides and angle 60° in-between), hence

|CD’ | = |D’E| = |CE|

Now triangle ADE is isosceles: |AE| = |DE| (each of them red + blue)

and angle AED = 120°, hence

angle(DAE) = angle(ADE) = 30°, finally angle(BAD) = 85° and angle(ADC) = 35°

P.S. (Having seen Mr. Daftary’s solution) Agree, it is a wonderful problem and Your solution, using symmetry, is also wonderful – I like it better than mine, since it is “look and see” solution.

Symmetry is another name of the Beauty in Geometry.

P.S.(2) Excellent solutions by Mr. Daftary and his son inspired me to make a second try – follow the next link:

http://farm3.static.flickr.com/2753/4319345365_189…

Reflect symmetrically ABCD with respect to the angle bisector of angle ABC. A and C are mutually symmetrical, let D’ is symmetric image of D, take on the angle bisector point E, such that angle CDE = 70°. Triangles ABD’ and CBD are congruent and since their acute angles are 5° the triangle BDD’ is isosceles with angle 60°, i.e. equilateral. Hence the 3rd yellow triangle EDD’ is congruent to both above, what implies

|DE| = |D’E| = |AB| etc.

Hence we have an equilateral hexagon ABCDED’ – a union of 6 overlapping identical copies of the initial quadrilateral ABCD – one of them is BCDE with angles 35°,170°, 70° and 85°. Its symmetry is shown on the right picture – 3 axial symmetries and 3 rotations at 0°, 120° and 240° – they comprise the dihedral group D3 of equilateral triangle’s symmetries.

1

http://i299.photobucket.com/albums/mm286/rozeta53/…

So, the intersection of the diagonals, the point O, is the circumcenter of the triangle ADE.

Edit1:

Beautiful symmetric solution of Mr. Daftary!

The second solution is nice, too.

Edit2:

Nice new solutions from Duke and Rakesh!

1

angle BAD = 85

170/2 = 85 you need to draw a parallelogram and you will see more details ..

Source(s): math professor in harvard

the angle is 80.9° = arccos ( 1- cos 70°-cos 60°)…but no calculator ??