Isosceles Circumcircle

$\bigtriangleup ABC$ is an isosceles triangle so $BD \perp AC$ .

We have $\bigtriangleup BED \sim \bigtriangleup DEC$ so $\frac {BD}{DC}=\frac {ED}{EC}$ .

Consider $\bigtriangleup DFB$ and $\bigtriangleup CEA$ :

We have that $\widehat{FDB} =90 ^ \circ$ - $\widehat{DBC} =\widehat{ECA}$ ,

and $\frac {DB}{CA}=\frac {DB}{2DC}=\frac {ED}{2EC}=\frac {2FD}{2EC}=\frac {DF}{CE}$

Thus, by SAS, $\bigtriangleup DFB \sim \bigtriangleup CEA$ . This implies that $\widehat{DAG}=\widehat{DBG}$

Hence $BGDA$ is a cyclic quadrilateral, so that $\widehat{BGA}=\widehat{BDA}=90 ^ \circ$ .

It turns out that the angle condition on ABC is irrelevant. We can solve this problem using cartesian coordinates:

Set A = (-b,0), B = (0,a), C = (b,0), D = (0,0). Hence E = (r,s) and F = (r/2, s/2), where ar + bs = ab (E lies on BC) and as = br (DE perpendicular to BC).

This gives that $r = \frac{a^2b}{a^2 + b^2}$ and $s = \frac{ab^2}{a^2 + b^2}$ .

Finally, we can easily evaluate the ratio $\frac{s-2a}{r} \frac{s}{r+b}$ , substituting values from above to obtain a value of -1, showing that BF and AE are perpendicular.

Hence, the answer is 90.

first of all, note that, the problem is independent on \angle ABC CONSTRUCTION:- draw BE' on BC, such that BE'=DE; draw C'E', perpendicular to BE' and the point C' lies on the line segment BD. Let, M is the mid point of EC, then, join D and M. Let C'E' and BF intersects each other at the point M' PROOF:- \Delta BAC is isosceles, so, BD\perp\AC where, D is the mid point of AC, and DE\perp\BC, so \angle C'BE' = \angle EDC \Delta BC'E' ≅ \Delta DCE [BE' = DE, \angle C'BE' = \angle EDC, \angle BE'C' = \angle DEC = 90 ^ \circ] and \Delta BM'E' ≅ \Delta DEM[M' is the mid point of C'E' by \Delta BC'E' \sim\ \Delta BDE] so, \angle M'BE' = \angle MDE and AE\parallel\DM so, \angle AED = \angle MDE[DM is the line segment, joining the mid points of the sides AC and EC of \Delta CEA] then, \angle AED = \angle MDE = \angle M'BE' = \angle GBE \angle BEG = 90 ^ \circ - \angle AED = 90 ^ \circ - \angle GBE so, \angle BGA = \angle GBE + \angle BEG = \angle GBE + 90 ^ \circ - \angle GBE so, \angle BGA = 90 ^ \circ[ANSWER]

Let $EBG=B_1$ , $GBD=B_2$ , $DAE=A_1$ and $DEA=E_1$ . We know that $EF=FD$ so two triangles $BEF$ and $BFD$ have equal area, it gives $\frac{1}{2}BE.BF.\sin B_1=\frac{1}{2}BE.BD.\sin B_2$ $\frac{\sin B_2}{\sin B_1}=\frac{BE}{BD}=\sin EDB$ its immediate that $EDB=C$ finally $\frac{\sin B_2}{\sin B_1}=\sin C \ \ \ (I)$ Now we use law of sines in triangle $ADE$ : $\frac{AD}{\sin E_1}=\frac{DE}{\sin A_1}$ $\frac{\sin A_1}{\sin E_1}=\frac{DE}{AD}=\frac{DE}{DC}=\sin C$ it gives: $\frac{\sin A_1}{\sin E_1}=\sin C \ \ \ (II)$ from two equations $(I)$ and $(II)$ we have $\frac{\sin B_2}{\sin B_1}=\frac{\sin A_1}{\sin E_1} \ \ \ (\star)$ notice that $A_1+E_1=\pi-ADE=\pi-(\frac{\pi}{2}+EDB)=\frac{\pi}{2}-C$ $A_1=\frac{\pi}{2}-C-E_1 \ \ \ (III)$ $B_1+B_2=\pi-(BED+EDB)=\pi-(\frac{\pi}{2}+EDB)=\frac{\pi}{2}-C$ $B_2=\frac{\pi}{2}-C-B_1 \ \ \ (IV)$ putting $(III)$ and $(IV)$ in $(\star)$ we get $\frac{\sin (\frac{\pi}{2}-C-B_1)}{\sin B_1}=\frac{\sin (\frac{\pi}{2}-C-E_1)}{\sin E_1}$ from here its easy to see that $E_1=B_1$ . In right triangle $BFE$ we have $BFE=\frac{\pi}{2}-B_1$ , Now we can write in triangle $FGE$ : $BGA=FGE=\pi-BFE-E_1=\pi-(\frac{\pi}{2}-B_1)-E_1=\frac{\pi}{2}$

Let $H$ be the midpoint of $AB$ . Since $BD\perp AC$ , hence $H$ is the circumcenter of $BAD$ , which we will call $\Gamma$ . Since $DH \parallel BC \perp DE$ , this shows that $DE$ is tangential to $\Gamma$ . Let $BF$ intersect $\Gamma$ at $G'$ .

Then, $FE^2 = FD^2 = FG' \cdot FB$ by power of a point. This shows that $FEG'$ and $FBE$ are similar triangles, and thus $FG'E$ is a right angle. Since $AG'B$ is also a right angle, it shows that $A, G', E$ are collinear, and thus that $G'$ is the intersection of $BF$ and $AE$ , and thus $G=G'$ . Hence, $\angle AGB = 90 ^\circ$ .

Let the distance $DC = m, m > 0$

And construct the triangle ABC on the Cartesian plane, with $A = (-m, 0)$ , $D$ be the origin, $C = (m,0)$

Since $\angle BCA = \frac {180-123}{2} = 28.5 ^ \circ$ , then the height of the triangle $DB = m \cdot \tan (28.5 ^ \circ )$

From here, I will denote $p = 28.5 ^ \circ$ for simplicity sake

So, $B = (0, m \cdot \tan p )$

The equation of $BC$ is $\frac {y-0} {x-m} = \tan (180 ^ \circ - p )$ or

$y = - \tan p \cdot (x - m)$

Since $BC$ is perpendicular to $DE$ , the product of their gradient is $-1$ , thus gradient of $DE = - \frac {1} {- \tan p } = \cot p$ , and therefore the equation of DE is $y = \cot p \cdot x$

When comparing equation $BC$ and equation $DE$ , we can solve for coordinate E:

$- \tan p \cdot (x - m) = x \cdot \cot p$

$x (- \tan p - \cot p ) = -m \cdot \tan p$

$\Large x = \frac { m \cdot \tan p } { \tan p + \cot p }$

$\Large x = \frac { m \tan^2 p } { \tan^2 p + 1 }$

$\Large x = \frac { m \tan^2 p } { \sec^2 p }$

$x = m \cdot \sin^2 p \Rightarrow y = m \cdot \sin^2 p \cdot \cot p = m \cdot \sin p \cos p$

$\Rightarrow E = ( m \cdot \sin^2 p , m \cdot \sin p \cos p )$

Since $F$ is the midpoint of DE, $\Large F = ( \frac { m \cdot \sin^2 p } {2} , \frac {m \cdot \sin p \cos p } {2} )$

So the gradient of $BF$ is

$\Large m_{BF} = \frac { m \cdot tanp - \frac { m \cdot sinp \cdot cosp } {2} } { \frac { m \cdot sin^2 p } {2} }$

$\Large m_{BF} = \frac { \sin p \cdot \cos^2 p - 2 \sin p } { \sin^2 p \cdot \cos p }$

And since $A = (-m, 0)$ , we can find the gradient of AE as well

$\Large m_{AE} = \frac { m \cdot \sin p \cdot \cos p } { m \cdot \sin^2 p + m }$

$\Large m_{AE} = \frac { \sin p \cdot \cos p } { \sin^2 p + 1 }$

And hence,

$\Large m_{AE} \cdot m_{BF} = \frac { \sin p \cdot \cos p } { \sin^2 p + 1 } \cdot \frac { \sin p \cdot \cos^2 p - 2 \sin p } { \sin^2 p \cdot \cos p }$

$\Large m_{AE} \cdot m_{BF} = \frac { \cos^2 p - 2 } { \sin^2 p + 1 }$

$\Large m_{AE} \cdot m_{BF} = \frac { \cos^2 p - 2 } { 2 - \cos^2 p }$

$\Large m_{AE} \cdot m_{BF} = -1$

Since their product of gradient is $-1$ , straight lines $BF$ and $AE$ must be perpendicular to one another.

Thus, $\angle BGA = 90 ^ \circ$

Draw the figure, then take D(0,0), A(a,0), B(0,b) & C(-a,0). As angle ABC=123, angle BAC & BCA = 28.5 AB=root(a^2+b^2) DB=b=root(a^2+b^2)sin(28.5) & b^2=(a^2+b^2)sin^2(28.5) From this equation we get: b=a tan(28.5) Thus B(0,a tan(28.5) and C(-a,0) From this we can get the equation of BC which is: y=tan(28.5)x+a tan(28.5) As DE is perpendicular to BC and passes through D(0,0) The equation of DE is: y = -cot(28.5)x From the equations of DE and BC we find the co-ordinate of E: E(-a sin^2(28.5), a sin(28.5)cos(28.5)) As F is the mid-point of DE, its co-ordinate is: F(-a/2 sin^2(28.5), a/2 sin(28.5)cos(28.5)) Using the co-ordinates of A(a,0) and E(-a sin^2(28.5), a sin(28.5)cos(28.5)) we find the slope of AE: (-sin(28.5)cos(28.5))/(1+sin^2(28.5)) Using co-ordinate of B(0, a tan(28.5)) and F(-a/2 sin^2(28.5), a/2 sin(28.5)cos(28.5)) we find the slope of BF: (-cos^2(28.5) - 2)/( sin(28.5)cos(28.5)) Multiplying the slopes we get: (cos^2(28.5) - 2)/(1+sin^2(28.5)) =(cos^2(28.5) - sin^(28.5) - cos^2(28.5) -1)/(1+sin^2(28.5)) =((-1) (sin^(28.5) + 1))/(1+sin^2(28.5)) = -1 Thus, BF and AE are mutually perpendicular to each other And therefore angle BGA must be 90 degrees

In the triangle ABC, let connect BD. BD is the median of the Isosceles triangle ABC. Now the median of an Isosceles Triangle is also the angle bisector , hence it bisects the angle ABC. BD is also the perpendicular bisector of the side AC.

As BD bisects angle ABC , therefore <CBD=61.5 deg

Now the desired <BGA= <EBF+ <AEB. -------------- (1)

as DE is perpendicular on BC , hence <BED=90

Now lets consider triangle ADE .

<AED= 90- <AEB and <EAD= <CAB-(180 -123 -<AEB)

Now as the triangle ABC is isosceles with AB=BC , hence <CAB=<BCA= (180-123)/2= 28.5

hence <EAD= 28.5 - (57 - <AEB)= <AEB - 28.5

So By applying the Sine rule in triangle ADE , we get

sin(90 - <AEB) / AD = sin (<AEB - 28.5)/ DE

sin(90 - <AEB)/ sin (<AEB - 28.5) = AD/DE= CD/DE ( As D is the midpoint of AC, hence AD=CD) ------------------------- (2)

Now from the right angled triangle CDE we can have

CD/DE= 1/sin 28.5 .

Hence denoting <AEB=x, the equation (2) can be written as

sin(90 - x) / sin(x-28.5) = 1/sin 28.5

now sin(x-28.5) = sin (-90+61.5+x) = - cos(61.5+x)

hence cos(61.5+x)/cos(x)= -cos(61.5) ( using the fact that sin(90-x)=cos x and sin (28.5) = cos (61.5) )

now by expanding cos(61.5+x) and dividing it by cos(x) we will get

2 cos (61.5)= sin (61.5) tan (x)

hence tan (<AEB) = 2/ tan (61.5) ------------------ (3)

Now lets move on to the right angled triangles BFE and BED

From the right angled triangle BED we can have tan (61.5) = DE/BE

similarly from the right angled triangle BEF we can have
tan (<EBF) =EF/BE

now as F is the midpoint of DE , hence EF= DE/2

tan (<EBF) = DE/2BE = 1/2 (DE/BE) = 1/2 tan (61.5) ---------- (4)

from equation 3 and 4 we have

tan (<AEB) = 2/tan(61.5)

tan (<EBF) = 1/2 tan (61.5)

Now from Equation 1 we have

<BGA= <EBF+ <AEB

so tan (<BGA) = tan (<EBF+ <AEB) = tan (< EBF) + tan (< AEB) / (1- tan (< EBF) tan (<AEB) )

now 1- tan (< EBF) tan (<AEB) = 1- (2 cot (61.5) * 1/2 tan (61.5) ) = 0

Therefore, tan (<BGA) tends to infinity hence <BGA=90.

AE perpendicular to BF