Just Some Isosceles Triangles

$\triangle ABC$ is an isosceles, hence $y = \angle BAC = \angle BCA$ .

Looking at $\triangle ABG$ which is a Right Angle triangle, $\angle BGA = 90-y$

By Opposite Vertical Angles, $\angle CGF = \angle BGA = 90 -y$

$\triangle CDE$ is an isosceles, hence $x = \angle DCE = \angle DEC$ .

Looking at $\triangle EDF$ which is a Right Angle triangle, $\angle DFE= 90-x$

By Opposite Vertical Angles, $\angle CFG= \angle DFE = 90 - x$

And now, $\angle BCD = 90^{\circ} = x + y + \angle GCF$

Thus, $\angle GCF = 90 - (x+y)$

Angles in a triangle add up to 180 in $\triangle CGF$

So, $\angle CGF + \angle CFG + \angle GCF = 180$ $(90-y) + (90-x) + (90-(x+y)) = 180$

Simplifies into, $x + y = 45^{\circ}$

Since $\angle GCF = 90 - (x+y) = 90 - 45 = 45^{\circ}$ $\boxed{\angle GCF = 45^{\circ}}$

This Question cant be SOLVED without applying EXTRA IQ or some INTELLIGENCE , including basic knowledge , concepts of Triangle . I was able to solve it bcause my triangle concepts are clear and VERY IMPORTANT , _ I thought to create rlationship between what is asked or to find out WITH what is GIVEN in the question / fig . In these finding out or proving questions.... Always write down the _ GIVEN PART _ for clear understanding . _ GIVEN _ :- _ AB=BC DE=DC ANGLE ABD , EDB , BCD = 90 _ TO find __ :- _ ANGLE GCF

SOLUTION :- Now lets apply our BASIC concepts of triangle ........ lets start with our given part.....

            In triangle ABC ,   __ AB=BC ( given )
                  and we know that angle opposite to the sides of isoscless triangle are equal ,
          thus , Angle BAC= BCA --------------- 1
             Similarly , Angle DEC= DCE ------------------2

NOW , consider triangle ABG , ABG + AGB + BAG = 180 ( U KNOW IT !!!! ) BUT ABG= 90 ........... MEANS , AGB + BAG = 90 __ we can write AGB = FGC ( opposite angles of two intersecting lines ) thus , FGC + BAC = 90 FGC + BCG = 90 ( from 1 )-------------3

Similarly , DCE + GFC = 90 ---------------4

NOW , we know GCF + CFG + CGF = 180 GCF + ( 90 - BCG ) + ( 90 + DCF ) = 180 GCF + 180 - ( BCG + DCF ) 180 GCF - ( BCG + DCF ) = 0-----------------5 bUT WE KNOW THAT , BCG + GCF + DCF = 90 BCG + DCF = 90 - GCF ------- 6

PUTTING THE 6 IN 5 .........

                          GCF + ( 90 - GCF ) = 0 
                           2 GCF = 90

_ THERFORE !!!! GCF = 45 !!!!!! _

There are (at least) a couple other ways of getting at $\ \angle CGF \$ ; I did three as a check, one of which is the method Mohammad Al Ali used. I will continue his notation.

One is that the "third angles" in the isosceles triangles have measure $\ m(\angle ABC) \ = \ 180^\circ \ - \ 2y \$ and $\ m(\angle EDC ) \ = \ 180^\circ \ - \ 2x \$ . Since $\ \angle ABG \$ and $\ \angle EDF \$ are right angles, so $\ m(\angle GBC) \ = \ 90^\circ \ - \ 2y \$ and $\ m(\angle FDC ) \ = \ 90^\circ \ - \ 2x \$ . Therefore, as the "third angle" of $\ \Delta BCD \$ , we must have

$\ m(\angle BCD) \ = \ 180^\circ \ - \ ( \ 90^\circ \ - \ 2y \ ) \ - \ ( \ 90^\circ \ - \ 2x \ ) \ = \ 2x \ + \ 2y \$ .

But we are given that $\ \angle BCD \$ is a right angle, so $\ \ 2x \ + \ 2y \ = \ 90^\circ$ . Angle BCD includes an angle from each of the isosceles triangles, so $\ m(\angle BCD) \ = \ \ m(\angle BCG) \ + \ m(\angle CGF) \ + \ m(\angle DCF) \ = \ y \ + \ m(\angle CGF) \ + \ x \$ . We have already established that $\ x \ + \ y \ = \ 45^\circ \$ , hence

$m(\angle CGF) \ = \ 90^\circ \ - \ (x + y) \ = \ 45^\circ \$ .

Another proceeds from the results Mohammad Al Ali finds for triangle GCF. The apex angle $\ \angle GCF \$ has measure

$180^\circ \ - \ m(\angle CGF) \ - \ m(\angle CFG) \ = \ 180^\circ \ - \ ( \ 90^\circ \ - \ y \ ) \ - \ ( \ 90^\circ \ - \ x \ ) \ = \ x \ + \ y \$ .

Together with $\ m(\angle BCD) \ = \ \ m(\angle BCG) \ + \ m(\angle CGF) \ + \ m(\angle DCF) \ = \ 90^\circ \$ , we have

$90^\circ \ = \ y \ + \ ( \ x \ + \ y \ ) \ + \ x \ = \ 2x \ + \ 2y \$ .

Thus, we again find $m(\angle CGF) \ = \ x \ + \ y \ = \ 45^\circ \$ .

Since <BCD is a 90 degree angle, Point C must be located along a circle with diameter endpoints B and D. Therefore <GCF (which is equal to <ACE) is a constant. The 2 extremes (max and min) of <ACE are when C and B are the same point and when C and D are the same point. In both cases, the angle of <ACE is 45 degrees.

because here A & E are right angle so if those are divided equally to a common point so its simply 45'

Solved by observation

Since the relative sizes of the isosceles triangles are irrelevant, redraw the diagram with $\overline{CD}$ and $\overline{DE}$ very tiny, approaching zero. Don't forget to keep $\angle C = 90 ^ \circ$ . Then $\triangle{ABC}$ approaches an isosceles right triangle and you can see that the desired angle is 45 degrees.