Geo question about lengths

Question

I have a question from a maths paper.

Let $ABC$ be an isosceles triangle with base $BC$ and Angle $BAC$ = 100. The bisector of Angle $ABC$ intersects $AC$ in $P$ . Show that $BC$ = $AP$ +$PB$.

Could I have some help solving this? Thanks

Answer 1

Don't know if coordiate method is allowed, let $A(0,\cos(50^\circ)), C(\sin(50^\circ),0), B(-\sin(50^\circ),0)$.
Bissector of $\angle ABC$ intersects $OA$ in $(0,\tan(20^\circ)\sin(50^\circ))$ so let's say $BP$ is $\{uB+(1-u)(0,\tan(20^\circ)\sin(50^\circ)),u\in\mathbb{R}\}$, $AC$ is $\{vA+(1-v)C,v\in\mathbb{R}\}$, $$P=AC\cap BP:u(-\sin(50^\circ),0)+(1-u)(0,\tan(20^\circ)\sin(50^\circ))= v(0,\cos(50^\circ))+(1-v)(\sin(50^\circ),0) $$ $$\begin{cases} x: -u\sin(50^\circ)=(1-v)\sin(50^\circ),\\ y: (1-u)\tan(20^\circ)\sin(50^\circ)=v\cos(50^\circ) \end{cases}$$ $$\begin{cases} u=v-1,\\ (1-(v-1))\tan(20^\circ)\cos(40^\circ)=v\sin(40^\circ) \end{cases}$$ $$\begin{cases} u=v-1,\\ (2-v)=2v\cot(20^\circ)\tan(40^\circ) \end{cases}$$ $$\begin{cases} u=v-1,\\ 2=v(2\cot(20^\circ)\tan(40^\circ)+1) \end{cases}$$ $|AP|=|P-A|=|vA+(1-v)C-A|=|1-v|\cdot|A-C|=|1-v|$,
$|BP|=|P-B|= |(u-1)B+(1-u)(0,\tan(20^\circ)\sin(50^\circ))|= |1-u|\cdot|(0,\tan(20^\circ)\sin(50^\circ))-B|= |1-u|\sqrt{\sin^2(50^\circ)+\tan^2(20^\circ)\sin^2(50^\circ)}$
We need to show that $|AP|+|BP|=2\sin(50^\circ)$ from here (we can wolframalpha this thing, see "x" result).

Maybe you could try (another way) some auxilary constructions like some line symmetries or so.
Or maybe recall any properties of angles $40^\circ$, $20^\circ$, $20^\circ$. Or use the sine rule to prove $\frac{PA}{PC}=\frac{BA}{BC}$, then find $PA$ from there and use a sine or cosine rule on $\Delta PAB$ find $PB$.

Answer 2

Construct isosceles triangle BDC and the congruent triangles ABP $\approx$ QBP. Then, the triangles PDC and PQC are congruent, which leads to PD + PQ = AP. Then,

$$BC = BD = BP + PD = BP + AP$$

Answer 3

Please take a look at the Pic.