# Solving for the value of $\angle CEB$ - $\frac{1}{4}$ $\angle CBA$ where $E$ is an exterior point of $\triangle ABC$

From point $$A$$ of $$\triangle ABC$$, a line $$AD$$ parallel to $$CB$$ is drawn so that $$AD=AB$$. From point $$B$$, a line parallel to $$AC$$ is drawn so that $$BE=BC$$. Point $$D$$ and $$E$$ lie on different sides of $$BC$$. If $$D$$, $$B$$ and $$E$$ are collinear, what is the value of $$\angle CEB$$ - $$\frac{1}{4}\angle CBA$$? My Attempt:

Let denote the $$\angle CEB$$ = $$\alpha$$ and $$\angle ABD$$ = $$\gamma$$

As $$BE = BC$$, so we can write that $$\angle BCE$$ = $$\alpha$$ also.

Likewise, $$\angle ADB$$ = $$\gamma$$ (as $$AB$$ = $$AD$$). $$AC$$ $$\parallel$$ $$BE$$ and $$BC$$ is their secant line, so $$\angle EBC$$ = $$\angle ACB$$ = $$\theta$$ (By denoting the angle as $$\theta$$)

Again, $$CB$$ $$\parallel$$ $$AD$$ and $$AB$$ is its secant line, so $$\angle CBA$$ =$$\angle BAD$$ = $$\beta$$

As, $$D$$, $$B$$ and $$E$$ are collinear, so $$\beta$$ = 180$$^\circ$$ - ($$\gamma + \theta$$).....(1)

Again, from $$\triangle ABD$$, $$\beta$$ can be written as $$\beta$$ = 180$$^\circ$$ - 2$$\gamma$$.....(2)

So, we got the equation from (1) and (2) that

180$$^\circ$$ - 2$$\gamma$$ = 180$$^\circ$$ - ($$\gamma + \theta$$) $$\implies$$ $$\gamma$$ = $$\theta$$

So, replacing $$\theta$$ by $$\gamma$$, $$\angle EBC$$ =$$\angle ACB$$ = $$\gamma$$

And, here I got stuck. I couldn't find a way out to proceed and solve the problem. I was unable to measure the value of $$\angle BAC$$. So, I couldn't make 3 equation and find the specific value of each angle.

• Maybe given that $AB=AD$? Feb 5, 2019 at 13:06 I don’t see why from $$AC∥BE$$, we get $$∠CBA = (∠CBE) =\alpha$$.
The way to do it is $$\gamma = \angle ABD = \angle D = \angle ACB$$ because ADBC is a //gm.
Let $$\angle ABC = \theta$$. From interior angle sums, we have $$2 \gamma + \theta = 180^0$$ and $$\gamma + 2 \alpha = 180^0$$
Result follows by eliminating $$\gamma$$.
• @AnirbanNiloy No! $\angle ECB = \angle CBA$ is true ONLY WHEN CE // AB.