# hard question. need geometric proof.

• Jan 28th 2011, 12:08 AM
abhishekkgp
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.
• Jan 28th 2011, 05:19 AM
ragnar
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.
• Jan 28th 2011, 05:46 AM
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?
• Jan 28th 2011, 05:48 AM
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

• Jan 28th 2011, 05:59 AM
ragnar
Ah, right, good catch.
• Jan 28th 2011, 07:09 AM
emakarov
Quote:

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?
• Jan 28th 2011, 07:27 AM
took $\angleNDC=40+x$ -totally idiotic
from $\Delta CDN$ you'll find $\angle MND=(80+x)^{\circ}$