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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.

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  • $\begingroup$ Maybe given that $AB=AD$? $\endgroup$ – Michael Rozenberg Feb 5 at 13:06
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enter image description hereI 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 $\gamma$.

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    $\begingroup$ @AnirbanNiloy No! $\angle ECB = \angle CBA$ is true ONLY WHEN CE // AB. $\endgroup$ – Mick Feb 5 at 18:20

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