Lines from a Triangle

By angle chasing, $\angle AFB =360^\circ -\angle FAC-\angle ACE-\angle CEB=180^\circ -\angle ACE= \angle ACD. \angle FAB =90-\angle BAC= \angle CAD$ . By SSS, triangles $FAB$ and $CAD$ are similar. Since $AB=AD$ , it follows that $FAB$ and $CAD$ are congruent, so $FA=CA$

Hence, $FCA$ is an isosceles right triangle, and $\angle FCB = 90^\circ - \angle FCA = 45^\circ$ .

$\angle BAD = \angle FAC = 90^\circ \Rightarrow \angle FAB = \angle CAD = 90^\circ - \angle BAC$ $\angle DAB = \angle DEB = 90^\circ \Rightarrow DAEB \text{ is cyclic}$ $\text{By angles in same segment, } \angle FBA = \angle CDA$ $\text{Also, } AB=AD \text{ as given.}$ $\text{Using AAS rule, Triangles FAB and CAD are congruent}$ $\text{Therefore, } FA=AC \Rightarrow \angle ACF = \angle AFC = 45^\circ$ $\angle FCB = \angle ACB - \angle ACF = 90^\circ - 45^\circ = 45^\circ$

I knew the answer could not change based on what triangle was used, so I used the easiest triangle, a right isosceles triangle. With AD being perpendicular and equal to AB, and then connecting D to C would essentially create another 45-45-90 triangle. After that, E is technically the same point as B, because DC passes through B. However, the line made by the perpendicular foot would still be vertical Because line DC is horizontal. Therefore By choosing, the point F that is on the same horizontal line as A, a right angle is formed by FAC. After this, it can clearly be seen that angle FCB is 45 degrees.

$\angle BAC + \angle CAD = \angle FAB + \angle BAC = 90^\circ$ , so $\angle CAD = \angle FAB$ . Also, $AD=AB$ . Let $G$ be the intersection of $AB$ and $DC$ .

If $\angle BAC > \angle ABC$ , then $G$ is inside $ACBF$ . $\angle BAD = \angle DEB = 90^\circ$ , and $\angle AGD = \angle EGB$ since they are vertical angles. Hence, $\angle ADG = \angle FBA$ .

If $\angle BAC < \angle ABC$ , then $G$ is outside $ACBF$ . $\angle BAD = \angle GEB = 90^\circ$ , and $\angle AGD = \angle EGB$ since they coincide. Hence, $\angle ADG = \angle EBG$ . $\angle EBG = \angle FBA$ because they are vertical angles. By transitivity, $\angle ADG = \angle FBA$ .

By ASA Congruence, $\bigtriangleup CAD \cong \bigtriangleup FAB$ , so $AF=AC$ . Thus, $\bigtriangleup ACF$ is an isosceles right triangle, and $\angle ACF = \angle FCB = 45^\circ$ .

• If $\angle BAC = \angle ABC$ , then $G$ coincides with $B$ . Here, ACBF is a square, and $\angle FCB = 45^\circ$ .

WE

Main thing here is being able to draw the triangles

Initially draw D is outside the triangle since BAD = 90 Next to draw E the only option is to Extend DC so that we can draw the perpendicular

Also note: Since AD = AB Angle ADB = ABC = 45

Next since FAC = 90 AF is parallel to CB Thus we have similar triangles ADB and FCB And each angles will be 45 degrees

We join $D,B$ and $A,E$ . Now, $\triangle DAB$ is a right angled isosceles triangle with $\angle ADB = \angle ABD = 45^{\circ}$ . Now we note the two cyclic quadrilaterals, namely, $AFEC$ and $ADEB$ . Now, we need to find $\angle FCB=\theta$ (say).We have $AF \parallel CB$ and hence

$\angle AFC = \theta = \angle AEC \text{[ since AFEC is cyclic ]} = \angle AED \text{ [since ABED is cyclic ]}= \angle ABD=45^{\circ}$

Hence, $\angle FCB=\boxed{45^{\circ}}$

As $\triangle DAB$ is an isosceles right triangle, $\angle ADB=90^\circ$ . Since $\angle DAB=\angle AEB=90^\circ$ , quadrilateral $DABE$ is cyclic, so $\angle ABE=\angle ADB=45^\circ$ . Next, since $\angle FAC$ is also a right angle, $FACE$ is a cyclic quad, so $\angle ACF=\angle AEF=45^\circ$ . Finally, we get that the desired answer is $\angle FCB=90^\circ-\angle ACF=\boxed{45^\circ}$ .