$ABC$ is an isosceles ($AB = AC$) and $\angle A = 100^{\circ}$. An point $D$ is on $AB$ so that $CD$ angle bisector $\angle ACB$. If $BC= 2017$ then calculate $AD+CD$.
The value of $AD+CD$ where $ABC$ is an isosceles triangle, $D$ bisects angle $ACB, BC = 2017$ unit
contest-matheuclidean-geometrygeometry
Related Solutions
$\triangle ABC \sim \triangle AMM' \sim \triangle PNM'$
$\frac{BC}{AB}=\frac{NM'}{PN}$
$PM'=PM=2PN$
$NM'=\sqrt{(2PN)^2-PN^2}=PN\sqrt3$
$\frac{NM'}{PN}=\sqrt3$
$\frac{BC}{AB}=\sqrt3$
$BC=AB\sqrt3=5\sqrt3$
$a=5,b=3,\color{blue}{a+b=8}$
This is an irregular hexagon so the sum of the interior angles is $720^\circ.$
$\angle p+\angle q +\angle r + \angle s + (360^\circ-\angle x) + (360^\circ - \angle y)= 720^\circ$
$30^\circ+45^\circ+50^\circ+25^\circ+360^\circ-\angle x+360^\circ-\angle y=720^\circ$
$150^\circ=\angle x+\angle y$
All the best for the olympiad.
Best Answer
Let $E$ be on $BC$ so $CE = CD$. Then $\angle DEC = 80^{\circ}$ so $CADE$ is cyclic, so we have by the power of the point with respect to $B$: $$a\cdot (a-d) = b\cdot (b-x)\;\;\;\;\;\;\;\;\;\;\;\;(1)$$
By angle bisector theorem we have $${b-x\over x} = {a\over b}\implies b(b-x)= ax\;\;\;\;\;\;\;\;(2)$$
By (1) and (2) we have $$a(a-d)= ax \implies a-d= x$$
Since we are searching for $d+x = a= 2017$ we are done.