euclidean-geometrygeometry
Let $\triangle {ABC}$ s.t. $AB=AC$ and $\angle{A}>90$. If $D\in [AB$ s.t. $AD=BC$ and $\angle{ADC}=\frac{3}{4}\cdot \angle{ABC}$ find $\angle {A}$.
My idea: I denote $\angle B=4x$, then I apply "Sine theorem" in $\triangle {ABC}$ and $ \triangle {ADC}$.
I obtain $sin(5x)=2sin(3x)\cdot cos(4x)$. Now I am stuck.
I try to construct and to solve it with an elementary construction but I didn't succeed. Can I apply here "The pants theorem"?
You can make the following.
By the law of sines get $AB$ and $AQ$, which gives $AP$ and $AQ$ and by the law of cosines get $PQ$ and $PN$.
Now, by law of cosines get $\measuredangle APQ$ and from this get $\measuredangle BPN.$
Also, by law of cosines get $BN$, by law of cosines again get $\measuredangle BPN$, which gives $\measuredangle NBM.$
Now, in $\Delta NBM$ we have now: $BM$, $BN$ and $\measuredangle NBM.$
Can you end it now?
$\angle ABC=\angle ABD+\angle DBC=80^\circ$.
\begin{align*} AB&=BC\\ \implies \angle CAB&=\angle BCA=(180^\circ-\angle ABC)/2=50^\circ. \end{align*}
Erect an equilateral triangle $ACE$ on base $AC$. Then $\triangle$s $ABE, CBE$ are congruent in opposite sense because $AB=CB$, $AE=CE$ and $BE$ is common. Thus $$\angle AEB=\angle BEC=30^\circ.$$
$$\angle CDB=180^\circ-\angle DBC-\angle BCD=150^\circ.$$ Thus quadrilateral $BDCE$ is cyclic because its angles $D$ and $E$ are supplementary. Thus $$\angle DEC=\angle DBC=20^\circ.$$
\begin{align*} \angle ECB&=\angle ECA-\angle BCA=10^\circ\\ \implies \angle ECD&=\angle ECB+\angle BCD=20^\circ=\angle DEC. \end{align*}
Thus triangle $CED$ is isosceles on base $CE$, so $CD=DE$. Thus $\triangle$s $ACD, AED$ are congruent in opposite sense because $AC=AE$, $CD=ED$ and $AD$ is common. Thus
\begin{align*} \angle CAD&=\angle DAE=30^\circ\\ \angle BAE&=\angle CAE-\angle CAB=10^\circ\\ \implies \angle DAB&=\angle DAE-\angle BAE=20^\circ\\ \implies \angle BDA&=180^\circ-\angle DAB-\angle ABD=100^\circ. \end{align*}
Best Answer
Let $E$ on $BC$ such that $CE\cong AC\cong AB$, and denote $\angle ABC = \alpha$.