Let $ABC$ be a triangle. Let $D$ and $E$ such that $AB \perp BD$, $AC \perp CE$Prove that $\bigtriangleup FBC$ is an isosceles right angled.

Question

PROBLEM:

Let $ABC$ be a triangle in which the measures $\angle ABC, \angle ACB$ are smaller than $45$. We consider that the points $D$ and $E$ such that $AB \perp BD$, $AB=BD$, $AC \perp CE$, $AC=CE$, and the points $D$ and $E$ are in the half-plane bounded by the line $BC$ which does not contain the point $A$. If $F$ is the midpoint of segment $DE$, prove that $\bigtriangleup FBC$ is an isosceles right angled.

MY DRAWING

Okey, so the first thing i thought of is that $\angle BAC>90$

Ee can clearly see that quadrilateral $ABQC$ can be inscribed in a circle, because $\angle ABQ+ \angle ACQ=180$. The center of this circle will be the middle of $AQ$.

I thought of making a rectangle that has the sides BF and FC and then show that it is actually a square.

I dont know what to do forward. Hope one of you can help me!

Answer 1

This is a synthetic proof using angle chasing mostly. We need to enhance OP’s sketch by introducing point $G$, which is the reflection of the point $A$ about the segment $BC$. We join $G$ to the five points $B$, $C$, $D$, $E$ and $F$ as shown in the diagram. For brevity, let $AB=c$ and $AC=b$, $\measuredangle ABC=\beta$, and $\measuredangle BCA=\omega$. Note that, by construction, $BD=AB=c$ and $CE=AC=b$.

Since $G$ is the reflection of $A$ about $BC$, we have, $$BG=BA=c\qquad\text{and}\qquad CG=CA=b. \tag{1}$$

Furthermore, $$\measuredangle CBG=\measuredangle ABC=\beta, \quad\text{and}\tag{2a}$$ $$\measuredangle GCB=\measuredangle BCA=\omega.\qquad\space \tag{2b}$$

It can be shown that $$\measuredangle BGC =\measuredangle CAB=\pi-\measuredangle ABC-\measuredangle BCA = \pi-\beta-\omega. \tag{3}$$

It is given that $\measuredangle ABD=\dfrac{\pi}{2}$. Therefore, using (2a),we shall write, that $$\measuredangle GBD=\measuredangle ABD-\measuredangle ABC-\measuredangle CAG=\dfrac{\pi}{2}-2\beta. \tag{4a}$$

In similar vein, using (2b), we can obtain, $$\measuredangle ECG =\dfrac{\pi}{2}-2\omega. \tag{4b}$$

According to (1), both triangles $GBD$ and $ECG$ are isosceles. Since we know the apex angles of both of them (see (4a} and (4b)), we can calculate their individual base angles, i.e., $\angle DGB$ and $\angle CGE$, as shown below. $$\measuredangle DGB=\dfrac{\pi-\measuredangle GBD }{2}=\dfrac{\pi}{4}+\beta\tag{5a}$$

Similarly, we can deduce, $$\measuredangle CGE=\dfrac{\pi}{4}+\omega.\tag{5b}$$

Using (3), (5a), and (5b), we can write, $$\measuredangle EGD=\pi - \measuredangle BGC - \measuredangle DGB - \measuredangle CGE =\dfrac{\pi}{2},$$ which means $\triangle EGD $ is a right-angled triangle and, therefore, the center of its circumcircle lies at $F$. This gives us $FG=FD=FE$. Hence, both quadrilaterals $BDFG$ and $CGFE$ are kites. Since the two diagonals of a kite perpendicularly bisect each other, $BF$ and $CF$ are the angle bisectors of $\measuredangle GBD$ and $\measuredangle ECG$ respectively. Therefore, according to (4a) and (4b), we have, $$\measuredangle CBF =\measuredangle CBG +\dfrac{1}{2}\measuredangle GBD = \dfrac{\pi}{4}, \quad\text{and}$$

$$\measuredangle FCB =\measuredangle GCB+\dfrac{1}{2}\measuredangle ECG = \dfrac{\pi}{4}.\qquad\space\space$$ Therefore, $\triangle BFC$ is not only isosceles, but also right-angled.

Answer 2

In the name of God
One of interesting ways to solve this is using coordinate geometry. draw Cartesian coordinate system in a way that x-axis crosses points B and C and y-axis crosses point A.
point A : (0,a)
point B : (-b,0)
point C :(c,0)
Note that a,b and c are bigger than zero.(X_A is always in the middle of X_B and X_C because ∠ABC,∠ACB are smaller than 45.)
Now we should calculate coordinates of points D and E in terms of a,b,c.
point D: (d,e)
point E : (f,g)
The gradient of line AB is equal to $\frac{a}{b}$ . So the gradient of line BD should be equal to $\frac{-b}{a}$ because AB is perpendicular to BD. We can calculate equation of line BD by having gradient($\frac{-b}{a}$) and a point (B).
y-Y_B = m(x-X_B)
y = $\frac{-b}{a}$(x+b)

e = $\frac{-b}{a}$(d+b)

We also know that AB = BD. As a result :
$\sqrt{a^2 + b^2}$ = $\sqrt{(d+b)^2 + e^2}$
By solving the system of equation for d,e, You will get:
d = a-b and e = -b
Or:
d= -a-b and e = b
Note that the second answer can't be correct. Because D is certainly at the right side of point B.
The gradient of line AC is also equal to $\frac{a}{-c}$. So the gradient of line CE is equal to $\frac{c}{a}$.
By doing the same calculations, We have :
y = $\frac{c}{a}$(x-c) => g = $\frac{c}{a}$(f-c)
And :
$\sqrt{a^2 + c^2}$ = $\sqrt{(f-c)^2 + g^2}$
The answers are :
g = -c , f = c-a
Or:
g = c , f = a+c
Again the second answer is not correct because Y_E should be negative.
coordinate of F, the middle point of D and E is : ($\frac{c-b}{2}$ , $\frac{-c-b}{2}$)
Now you can check the gradients of FC and FB :
m_FC = $\frac{\frac{-c-b}{2}}{\frac{c-b}{2} - c}$ = 1.
m_FB = $\frac{\frac{-c-b}{2}}{\frac{c-b}{2} + b}$ = -1
m_FC m_FB = -1 => They are perpendicular.
You can also check the distances :
$\sqrt{(\frac{-c-b}{2})^2 +(\frac{c-b}{2} - c )^2}$ = $\sqrt{(\frac{-c-b}{2})^2 +(\frac{c-b}{2} + b )^2}$ => FB=FC
So FBC is an isosceles right angled triangle.
Note that you could draw Cartesian coordinate system in any other places too, and still you'd be able to solve the question.