G1.3 In Figure $1, \triangle A B C$ is an isosceles triangle and $A B=A C$. Suppose the angle bisector of $\angle B$ meets $A C$ at $D$ and $B C=B D+A D$. Let $\angle A=c^{\circ}$, find the value of $c$.
Let $A B=n=A C ; A D=q, B D=p, C D=n-q$
$\angle A B D=x=\angle C B D ; \angle A C B=2 x$.
Let $E$ be a point on $B C$ such that $B E=p, E C=q$
Apply sine formula on $\triangle A B D$ and $\triangle B C D$.
$$
\frac{n}{\sin \angle A D B}=\frac{q}{\sin x} ; \frac{p+q}{\sin \angle B D C}=\frac{n-q}{\sin x}
$$
$\because \sin \angle A D B=\sin \angle B D C$
Dividing the above two equations
$$
\begin{array}{l}
\frac{n}{p+q}=\frac{q}{n-q} \\
\frac{A B}{B C}=\frac{E C}{C D} \text { and } \angle A B C=\angle E C D=2 x
\end{array}
$$
$\triangle A B C \sim \triangle E C D$ (2 sides proportional, included angle)
$\therefore \angle C D E=2 x$ (corr. $\angle \mathrm{s}, \sim \Delta$ 's)
$\angle B E D=4 x$ (ext. $\angle$ of $\triangle C D E)$
$\angle B D E=4 x(B D=B E=p$, base $\angle \mathrm{s}$, isos. $\triangle)$
$\angle A D B=3 x($ ext. $\angle$ of $\triangle B C D)$
$2 x+3 x+4 x=180^{\circ}$ (adj. $\angle \mathrm{s}$ on st. line $\left.A D C\right)$
$x=20^{\circ}$
$c^{\circ}=180^{\circ}-4 x=100^{\circ}(\angle \mathrm{s}$ sum of $\triangle A B C)$
Method 2
Claim $c^{\circ}>90^{\circ}$
Proof: Otherwise, either $\mathrm{c}^{\circ}90^{\circ}$
Locate a point $F$ on $B C$ so that $B F=n$
$\triangle A B D \cong \triangle F B D$ (S.A.S.)
$D F=q($ corr. sides $\cong \Delta \mathrm{s})$
$\angle D F B=c^{\circ}$ (corr. $\angle \mathrm{s} \cong \Delta \mathrm{s}$ )
$\angle D F C=180^{\circ}-c^{\circ} \angle 90^{\circ}$ (adj. $\angle$ s on st. line)
Locate a point $E$ on $F C$ so that $D E=q$
$\triangle D E F$ is isosceles
$\angle D E F=180^{\circ}-c^{\circ}$ (base $\angle \mathrm{s}$ isos. $\triangle \mathrm{s}$ )
$\angle D E C=c^{\circ}$ (adj. $\angle \mathrm{s}$ on st. line) ..... (3)
$\angle A B D=x=\angle C B D ; \angle A C B=2 x \ldots \ldots$ (4)
Consider $\triangle A B C$ and $\triangle E C D$
$\angle B A C=c^{\circ}=\angle C E D$ (by (3))
$\angle A B C=2 x=\angle E C D$ (by (4))
$\therefore \triangle A B C \sim \triangle E C D$ (equiangular)
$C E: E D=B A: A C$ (cor. sides, $\sim \Delta \mathrm{s}$ )
$C E: E D=1: 1(\because \triangle A B C$ is isosceles $)$
$\therefore C E=E D=q$
$B E=B C-C E=(p+q)-q=p$
$\because B E=p=B D$
$\therefore \triangle B D E$ is isosceles
$\angle B E D=\angle B D E=180^{\circ}-c^{\circ}$ (base $\angle$ s isos. $\triangle$ )
In $\triangle B D E, x+2\left(180^{\circ}-c^{\circ}\right)=180^{\circ}(\angle \mathrm{s} \mathrm{sum}$ of $\triangle) \Rightarrow x=2 c^{\circ}-180^{\circ}$. $\qquad$
In $\triangle A B C, c^{\circ}+4 x=180^{\circ}(\angle \mathrm{s}$ sum of $\triangle) \ldots \ldots$ (6)
Sub. (5) into (6), $c^{\circ}+4\left(2 c^{\circ}-180^{\circ}\right)=180^{\circ}$
$c=100$