Do you mean obtuse isosceles triangle $ABC$? Summary of below(so you can ignore this post): It didn't work Trigonometry So far, let angle $DBA$ be $x$, so $ACB=2x$ and $CAB=180-4x$. Then, labelling the rest of the angles, we get $ADB=3x$, and $CDB=180-3x$. Then, have side segment while $AD=a$ and $DC=b$, so $AB=a+b$. From the angle bisector theorem, we get $\frac{a+b}c=\frac{b}{a}$ where $c$ is side $BC$, meaning that $a(a+b)=bc$ and $c=\frac{a(a+b)}{b}$. Now, apply the Law of Cosines to triangle $ADB$, so $$BD^2=b^2+(a+b)^2-2\cos{(180-4x)}.$$THis simplifies to $$BD^2=2b^2+a^2+2ab+2\cos{4x}.$$ Next, we apply the Law of Cosines to triangle $CDB$, and using $DCB$ as the reference angle as $BD$'s length is still nasty and $2x$ is half of $4x$(maybe Double-Cosine Angles/something). This gets us $$BD^2=a^2+\left(\frac{a^2(a+b)^2}{b^2}\right)-2\cos{2x},$$finally simplifying down to $$BD^2=2a^2+\frac{a^4}{b^2}+\frac{2a}{b}-2\cos{2x}.$$ We set the two equations we got equal to each other, meaning that $$2b^2+a^2+2ab+2\cos{4x}=2a^2+\frac{a^4}{b^2}+\frac{2a}{b}-2\cos{2x}.$$Moving the cosines to one side, everything can semi-factor into $$a(a-\frac{a^3}{b}-\frac{a}{b})-2b(a+b)=2\cos{4x}+2\cos{2x}.$$ We can use the identity $\cos{2x}=2\cos^2{x}-1$, but it doesn't look so good from here.

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Yea proving that seems kinda sus.

My bad . $BC = BD + AD$ is not wanted but given. wanted is$ \angle A$ so the wording is not Quote: In an isosceles triangle $ABC$ ($AB = AC$), the bisector of the angle $B$ intersects $AC$ at point $D$. Prove that $BC = BD + AD$. but Quote: In an isosceles triangle $ABC$ ($AB = AC$), the bisector of the angle $B$ intersects $AC$ at point $D$ such that $BC = BD + AD$. Find $\angle A$. sorry