# Thread: hard question. need geometric proof.

1. ## hard question. need geometric proof.

ABC is an acute angled triangle. $\angle BCA=80 ^{\circ}$. The angle bisector of $\angle BCA$ meets AB in D. Let the perpendicular bisector of side AB meet the angle bisector of $\angle BDC$ in E. Join E and C. Its given that CE bisects exterior $\angle DCA$ (i.e. $\angle DCE=70^{\circ}$).
Prove that $\angle CBA=20^{\circ}$

The figure i have attached is not very clear but you do not need the figure. the text i have written is sufficient to understand the question.

2. I'm still working on this, but angle DCE is 90. Angle BCA is 80, so the exterior is 100, so each half is 50, and so on.

3. ragnar,
i think what abishek mentioning is that DC and the extending line is bisected by the line CE not BC and the extending line so i think conclusion- "each half is 50" is wrong,isn't it?

4. my solution,

if the $\^{B}=x^{\circ}$

let the point intersecting EF and BC be M and the point intersecting ED and BC be N.

since $\^{F} = 90^{\circ}$ $\angle NMF = (90+x)^{\circ}$

consider point D, since $\^{A}=(100-x)^{\circ}$ $\angle EDF=(70-\frac{x}{2})^{\circ}$

from $\Delta CDN$ you'll find $\angle MND=(80+x)^{\circ}$

now sum up all the angles in the quadrilateral and find x

5. Ah, right, good catch.

from $\Delta CDN$ you'll find $\angle MND=(80+x)^{\circ}$
Could you explain this a bit more? I see that $\angle MND$ is external to $\triangle CDN$, so $\angle MND=\angle NCD+\angle CDN=40+\angle NDF=40+70-x/2=110-x/2$. How do we get that it is $80+x$? Also, did you use the fact that AF = FB?

7. sorry, my mistake.

took $\angleNDC=40+x$ -totally idiotic

from $\Delta CDN$ you'll find $\angle MND=(80+x)^{\circ}$