In $$\triangle$$ABC, AD $$\perp$$ BC and BE $$\perp$$ AC. AD and BE intersect each other at F . If BF = AC, then the measure of $$\angle$$ABC is:

From the given question,

In $$\triangle BDF$$ and $$\triangle AEF$$

$$\angle BDF=\angle AEF=90$$

$$\angle BFD=\angle AFE$$  (opposite angle)

So, $$\angle DBF=\angle FAD$$

Hence $$\triangle BDF$$ and $$\triangle AEF$$ are similar triangle.

So, $$\triangle BDF$$ and $$\triangle ADC$$ are similar triangle

Hence, $$\dfrac{BF}{AC}=\dfrac{BD}{AD}$$

Hence, BD=AD

$$\tan \angle B=\dfrac{AD}{BD}=\tan 45^\circ$$
$$\angle ABX=45^\circ$$