This is the theorem to prove.
Below is my proof that I consider rather long and complex.
The given data is on this drawing:
- Construct $\angle DCE = \angle DCB$. The point $E$ on ray $CE$ is chosen in such way that $CE = CB$, and that is always possible by segment construction axiom.
- Connect points $B$, $E$, and $D$.
- $_\Delta CDB \cong _\Delta CDE$ by SAS, because $CD$ is their common side, $CB = CE$ by construction, $\angle DCB = \angle DCE$ by construction. Therefore $\angle DBC = \angle DEC$, as they are opposite to side $CD$ of these triangles.
- Let $O$ be the intersection point of $AD$ and $CE$. Also connect points $A$ and $C$ to construct a line $AC$.
- $_\Delta ABC$ is isosceles because $AB = BC$ as given by the statement of the theorem. Then $\angle BCA = \angle BCA$.
- Since $\angle BAD = \angle BCE$, then $\angle ECA = \angle DAC$ as well.
- We have $AO = CO$ because $\angle OAC = \angle OCA$ implies $_\Delta AOC$ is isosceles. As $AD = CE$, we get $AO + OD = CO + OE$. As $AO = CO$, we get $AO + OD = AO + OE$, which implies $OD = OE$, therefore $_\Delta OED$ is also isosceles. Even more, $\angle OED = \angle ODE = \angle OAC = \angle OCA$.
- $_\Delta AED \cong _\Delta CDE$ by SAS because $AD = CE$, $ED$ is their common side, and $\angle ADE = \angle CED$. This implies $\angle EAD = \angle ECD$, therefore $\angle EAB = 2\alpha – \alpha = \alpha$, which implies $AE$ is on the angle bisector of $\angle DAB$.
- $_\Delta DAB$ is isosceles since $AB = AD$ by theorem's statement. Therefore, angle bisector is also median and altitude. Then median and altitude of $_\Delta BED$ to side $BD$ also coincide, therefore $_\Delta BED$ is isosceles with $BE = ED$. Since also $BD = DE$, $_\Delta DEB$ is equilateral, therefore all internal angles of this triangle are equal to one another, $60^\circ each$.
- $_\Delta AEB \cong _\Delta AED$ by SAS because $AE$ is their common side, $AB = AD$, $\angle EAD = \angle EAB$. Therefore, $\angle ABE = \angle ADE$.
- As points $B$ and $O$ are both equidistant from points $A$ and $C$, they lie on perpendicular bisector of $AC$. Therefore, $BO$ is also median and angle bisector of $\angle ABC$. But same goes for $_\Delta BED$ and its angle $\angle EBD$. Then $\angle EBO = \angle DBO = 0.5\cdot 60^\circ = 30^\circ$.
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By using the notation on the last image, $\angle ABC = \beta + 30^\circ + 30^\circ + \beta = 60^\circ + 2\beta$.
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Consider the right triangle $_\Delta BCG$. Its acute angles sum up to $90^\circ$, therefore:
\begin{align*}
30^\circ + \beta + \alpha + \alpha + \beta &= 90^\circ \\
2\alpha + 2\beta &= 60^\circ \\
\alpha + \beta &= 30^\circ \\
\beta &= 30^\circ – \alpha
\end{align*} -
Now we are ready to express $\angle ABC$:
$$
60^\circ + 2\beta
= 60^\circ + 2(30^\circ – \alpha)
= 60^\circ + 60^\circ – 2\alpha
= 120^\circ – 2\alpha
$$
Which was to be proven.
What are shorter alternatives to that proof? I'm considering trigonometry, too, but I would prefer good old elementary geometry.
Best Answer
Here's a possible shorter proof. Reflect $A$ about $BD$ to $A'$. As $\angle DA'B=2\angle DCB$, then $A'$ is the center of the circle through $BDC$. Hence $BA'C$ is an equilateral triangle and $DA'C=60°-2\alpha$.
On the other hand $DBC={1\over2}DA'C=30°-\alpha$ and $ABD=90°-\alpha$, so that the requested value of $\angle ABC$ folows.