# hard question. need geometric proof.

• Jan 28th 2011, 12:08 AM
abhishekkgp
hard question. need geometric proof.
ABC is an acute angled triangle. $\displaystyle \angle BCA=80$^{\circ}$$. The angle bisector of \displaystyle \angle BCA meets AB in D. Let the perpendicular bisector of side AB meet the angle bisector of \displaystyle \angle BDC in E. Join E and C. Its given that CE bisects exterior \displaystyle \angle DCA (i.e. \displaystyle \angle DCE=70^{\circ}$$).
Prove that $\displaystyle \angle CBA=20$^{\circ}$$The figure i have attached is not very clear but you do not need the figure. the text i have written is sufficient to understand the question. • Jan 28th 2011, 05:19 AM ragnar I'm still working on this, but angle DCE is 90. Angle BCA is 80, so the exterior is 100, so each half is 50, and so on. • Jan 28th 2011, 05:46 AM BAdhi ragnar, i think what abishek mentioning is that DC and the extending line is bisected by the line CE not BC and the extending line so i think conclusion- "each half is 50" is wrong,isn't it? • Jan 28th 2011, 05:48 AM BAdhi my solution, if the \displaystyle \^{B}=x^{\circ}$$

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

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

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

from $\displaystyle \Delta CDN$ you'll find $\displaystyle \angle MND=(80+x)$^{\circ}$$now sum up all the angles in the quadrilateral and find x the answer will be 20 • Jan 28th 2011, 05:59 AM ragnar Ah, right, good catch. • Jan 28th 2011, 07:09 AM emakarov Quote: Originally Posted by BAdhi from \displaystyle \Delta CDN you'll find \displaystyle \angle MND=(80+x)^{\circ}$$

Could you explain this a bit more? I see that $\displaystyle \angle MND$ is external to $\displaystyle \triangle CDN$, so $\displaystyle \angle MND=\angle NCD+\angle CDN=40+\angle NDF=40+70-x/2=110-x/2$. How do we get that it is $\displaystyle 80+x$? Also, did you use the fact that AF = FB?
• Jan 28th 2011, 07:27 AM
took $\displaystyle \angleNDC=40+x$ -totally idiotic
from $\displaystyle \Delta CDN$ you'll find $\displaystyle \angle MND=(80+x)$^{\circ}