how to solve this problem? ( Euclid Geometric )

Question

Condition: For the quadrilateral $ABCD$, we have $BC=CD=DA$, and two angles are given: $\widehat{ADC}=96^\circ$, $\widehat{BCD}=48^\circ$.

Problem: Which is the measure of the angle $\widehat{ABC}$?

I can't solve this problem... I got an answer "162" by drawing it and measuring it, but I don't know how the answer came out Logically.

As I pondered it, I think it would be enough to show that the point $H$ is the circumcenter of $A'B'C$, but I could not prove it. How do you get $X$?

Answer 1

The angle of vertices in a pentagon is $108°$, this is the main key to solve this problem. After constructing a pentagon as shown in this diagram, we notice that the line segment $ED$ is a symmetry line of the pentagon and because $AD=BE$, then $\Delta ABF$ and $\Delta DEF$ are two similar isosceles triangles.

Thus: $\widehat {ABC} = 54°+108°=162°$.

Answer 2

HINT:

Calculate AC by SSA in triangle ADC. Reduce angle at C by the calculated amount.

As two sides and the non-included angle is given for the remaining triangle ABC we have two solutions by the ambiguous case.

Answer 3

Here is one more solution, which tries to use the idea to attack the problem as in the OP. In particular, i will use the points $C$ and $H$ from the second picture in the OP, they lead immediately to a solution. (This $C$ will be denoted below by $E$ to avoid the confusion with the point $C$ from the given quadrilateral. So $E$ is constructed so that $ADCE$ is a parallelogram. It turns out that $H$ is indeed the circumcenter of $\Delta BCE$, and the points $A,B,H$ are colinear.)

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Let us construct the point $E$, so that $ADCE$ is a parallelogram. Its angles in $D,E$ have measure $96^\circ$ each, its angles in $A,C$ have $84^\circ$ each.

We further construct $X$ in the interior of the parallelogram $ADCE$ (or in the half-plane containing $C,D$ w.r.t. the line $AE$) so that $\Delta AEX$ is equilateral.

Then the angles $\widehat {XEC}$, and respectively $\widehat {BCE}$ have measure $96^\circ-60^\circ=36^\circ$, and $86^\circ-48^\circ=36^\circ$. This information, joined with the equality (of lengths) of segments $BC=CE=EX$, implies that ($\Delta ECX=\Delta BCE$, that $BXCE$ is an isosceles trapezium with "easy angles", and finally that) these segments are three of the diagonals of a regular pentagon. Let $Y$ be the missing vertex of this pentagon, although not needed in the proof. Let $H$ be the circumcenter of the above regular pentagon.

It is time to display the picture so far:

Then the symmetries of the two marked regular polygons w.r.t. the common side $EX$ show that on the perpendicular bisector $(d)$ of $EX$ there are the points $B,H$ (pentagon symmetry) and $A$ (equilateral symmetry). From the collinearity of $A,B,H$, and since the angle built by $(d)$ with the direction $AD\|CE\|BX$ is $\widehat{HBX}=\frac 12\widehat{EBX}=\frac12 108^\circ=54^\circ$, we obtain the angle in $A$ in $ABCD$, and finally the angle in $B$, which is $360^\circ-(48^\circ+96^\circ+54^\circ)=\color{blue}{162^\circ}$, as claimed in the OP.

$\square$

Answer 4

I had proposed in my early comment that the question might be resolved using isosceles triangles (since the geometry seemed to suggest this rather strongly). A challenge arises, but not in the way I thought it would...

The two isosceles triangles that can be marked within the quadrilateral permit us to determine measures for all of the angles indicated. Two which cannot be found by this ploy alone will be identified as $ \ m(\angle BAC) = \theta \ $ and $ \ m(\angle ABD) = \phi \ \ . $ We are able to say at this point that $ \ \phi + \theta \ = \ 108º \ \ . $

Dividing the quadrilateral with $ \ \overline{BD} \ $ separates in into one of the isosceles triangles and another triangle. We will call the length of the congruent sides $ \ s \ \ , $ the length of $ \ \overline{BD} \ $ will be $ \ Y \ \ . $ and the length of the non-congruent side of the quadrilateral is $ \ X \ \ . $ The Law of Sines then permits us to write the relations

$$ \frac{\sin 66º}{s} \ = \ \frac{\sin 48º}{Y} \ \ \ \text{and} \ \ \ \frac{\sin \phi}{s} \ = \ \frac{\sin 30º}{X} \ = \ \frac{\sin (42º + \theta)}{Y} \ \ . $$

We do this also with the segment $ \ \overline{AC} \ $ to which we assign the length $ \ Z \ $ . From the Law of Sines, we have

$$ \frac{\sin 42º}{s} \ = \ \frac{\sin 96º}{Z} \ \ \ \text{and} \ \ \ \frac{\sin \theta}{s} \ = \ \frac{\sin 6º}{X} \ = \ \frac{\sin (66º + \phi)}{Z} \ \ . $$

We can now bring this ratios together to obtain

$$ X \ = \ \frac{\sin 30º}{\sin \phi} · s \ = \ \frac{\sin 6º}{\sin \theta} · s \ \ , \ \ Y \ = \ \frac{\sin 48º}{\sin 66º} · s \ = \ \frac{\sin (42º + \theta)}{\sin \phi} · s \ \ , $$ $$ Z \ = \ \frac{\sin 96º}{\sin 42º} · s \ = \ \frac{\sin (66º + \phi)}{\sin \theta} · s \ \ . $$

This is where the difficulty in this approach emerges, as we have a plethora of information with no clear guide as to how to make use of it. If all we cared about was getting an answer to the problem, we can easily produce expressions from which we can compute angles, such as

$$ \frac{\sin 96º}{\sin 42º} \ = \ \frac{2 \sin 48º \cos 48º}{\cos 48º} \ = \ 2 \sin 48º \ = \ \frac{\sin (66º + \phi)}{\sin \theta} $$ $$ \Rightarrow \ \ 2 \sin 48º \sin \theta \ = \ \sin 66º \cos \phi \ + \ \cos 66º \sin \phi \ \ , $$ which together with $$ \frac{\sin 30º}{\sin \phi} \ = \ \frac{\sin 6º}{\sin \theta} \ \ \Rightarrow \ \ \sin 30º · \sin \theta \ = \ \sin 6º · \sin \phi \ \ \Rightarrow \ \sin \theta \ = \ 2 \sin 6º · \sin \phi $$ $$ \Rightarrow \ \ 4 \sin 48º · \sin 6º · \sin \phi \ = \ \sin 66º \cos \phi \ + \ \cos 66º \sin \phi $$ $$ \Rightarrow \ \ \tan \phi \ = \ \frac{\sin 66ª}{4 \sin 48º · \sin 6º \ - \ \cos 66º} \ \ \Rightarrow \ \ \phi \ = \ -84º \ \ \text{or} \ \ 96º $$ very precisely by a calculator. We can achieve similar sorts of equations for $ \ \phi \ \ \text{or} \ \ \theta \ \ . $ (I had initially believed, because all of the marked angles are multiples of 6º or 12º, that double-angle and half-angle relations would be far more helpful than they turned out to be.)

But it would be more satisfying mathematically if we can obtain an exact result. The best option (after various attempts) seems to be simply to apply $ \ \sin \theta \ = \ 2 \sin 6º · \sin \phi \ $ and exploit its resemblance to the double-angle formula for sine, taking $ \ \sin \phi \ = \ \cos 6º \ $ and so $ \ \sin \theta \ = \ \sin ( 2 · 6º) \ \Rightarrow \ \theta \ = \ 12º \ \ . $ From that, we have $ \ \phi \ = \ 108º - \theta \ = \ 96º \ \ , $ which is consistent with $ \ \sin 96º \ = \ \cos 6º \ $ and with all the other ratio equations we established earlier.

Hence, we may conclude that $ \ m(\angle ABC) \ = \ 66º + \phi \ = \ 66º + 96º \ = \ 162º \ \ . $