Proving that a Quadrilateral is cyclic made by intersection of perpendiculars from $B,C,D$ from $\triangle ABC$ where $D$ is a pont on $BC$ .

Question

(sorry, couldn't come up with a better title)

The question:

$D$ is a point in the base $BC$ of $\triangle ABC$ and through $B, D, C$ lines are drawn perpendicular to $AB,AD,AC$ respectively meeting one another in $E,F,G$. Prove that $A,E,F,G$ are concyclic.

the fig:

In quadrilateral $AEBD$, $\angle EBA = \angle ADE = 90^{\circ}$. Thus quadrilateral $AEBF$ is cyclic. The center of the circle circumscribing this quadrilateral is the mid point of $AE$.

In quadrilateral $ADGC$, $\angle ADG = \angle ACG = 90^{\circ}$. Thus it is a cyclic quadrilateral. The center of the circle circumscribing this quadrilateral is the mid point of $AG$.

In quadrilateral $ABFC$, $\angle ABF= \angle ACF = 90^{\circ}$. Thus it is a quadrilateral. The center of the circle circumscribing this quadrilateral is the mid point of $AF$.

How to proceed from here?

Answer 1

Hint: Working backwards / wishful thinking.

If claim is true, then

1. $\angle EFG + \angle EAG = 180^\circ$.
2. OP identified $ABFC$ is cyclic, so $\angle BFC + \angle BAC = 180^\circ$.
3. 1 and 2 implies $ \angle EAG = \angle BAC$.
4. 3 implies $\angle EAB = \angle CAG$.
5. OP has identified other cyclic quads, which imply 4.

Now write up the solution in the forward direction.

Answer 2

We wish to show that $\angle EFG=180^{\circ}-\angle EAG$.
Define, $\angle EAB=\alpha,~ \angle BAD=\beta,~ \angle DAG=\gamma. $
Since $ADBE$ and $ACGD$ are cyclic, $$\angle CBF=\alpha+\beta ~~~\text{and}~~~ \angle DCG=\gamma.$$ Hence, in $\triangle BCF$, $$\angle BFC=\angle EFG=180^{\circ}-\alpha-\beta-\gamma.$$

Answer 3

In the picture, the angles marked $\alpha$ are equal because, as the OP has stated, $AEBD$ is a cyclic quadrilateral.

Likewise, the angles marked $\beta$ are also equal.

The complementary angle of $\alpha$, i.e. $90-\alpha$ is marked $\alpha’$, and likewise for $\beta$, using the data about perpendiculars given in the question.

Therefore, $$\angle EAG=\alpha’+\beta’=180-(\alpha+\beta)$$

Furthermore we see in $\triangle BCF$ that $$\angle BFC=180-\alpha’-\beta’=\alpha+\beta$$

Hence $\angle EAG+\angle BFC=180$, which proves that $AEFG$ is a cyclic quadrilateral.