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.
Best Answer
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 $\gamma$.