The problem is wrong and I do not see how to fix it. The only non-trivial right angle that I can see is $\angle BGD = 90^\circ$, but the triangle $\triangle BGD$ is already constructed. Anyway, the intersection H lies on the A-altitude of the triangle $\triangle ABC$ (see Another geometry question in the high school section for a proof). Let K be the foot of this A-altitude. Denote a = BC, b = CA, c = AB. Since K is the altitude foot and D the angle bisector foot, $BK = c \cos \widehat B,\ \ BD = \frac{ca}{b + c},\ \ BK \cdot BD = \frac{c^2a}{b + c} \cos \widehat B$ The power of the point B to the circumcircle (P) of the triangle $\triangle AFH$ is $BH \cdot BG = BA \cdot BF = BA \cdot BD \cos \widehat B = \frac{c^2a}{b + c} \cos \widehat B$ It follows that $BH \cdot BG = BK \cdot BD,\ \ \frac{BG}{BD} = \frac{BK}{BH}$ Thus the triangles $\triangle BKH \sim \triangle BGD$ having the common angle at the vertex B are similar by SAS, which means that the angle $\angle BGD = \angle BKH = 90^\circ$.

Attachments:

Quote: Prove that the triangle constructed from $BG$, $GE$ and $BF$ is right-angled. It should have been BG, GE, BF instead of BG, CE, BF. Problem corrected now.

Thanks, orl. The above solution then continues: Since $\angle DGB = \angle DGE = 90^\circ$ is right, $DG^2 = BD^2 - BG^2 = DE^2 - GE^2$ The right angle triangles $\triangle ADF \cong \triangle ADE$ are trivially congruent by ASA, because AD bisects the angle $\angle EAF \equiv \angle CAB$, hence DF = DE and $BD^2 - BG^2 = DF^2 - GE^2$ $BD^2 - DF^2 = BG^2 - GE^2$ Since $DF \perp AB$, the triangle $\triangle BDF$ is right and $BD^2 - DF^2 = BF^2$, so $BF^2 = BG^2 - GE^2$ $BF^2 + GE^2 = BG^2$ BG would be the hypotenuse of the right angle triangle constructed from BF, GE, BG.

I just wanted to add a synthetic proof of < BGD = 90° (that's what Yetti proved in post #2): At first, we apply the result of http://www.mathlinks.ro/Forum/viewtopic.php?t=41286 and http://www.mathlinks.ro/Forum/viewtopic.php?t=5460 to see that the point of intersection H of the lines BE and CF lies on the altitude of triangle ABC issuing from the vertex A, so that $AH\perp BC$. We will use directed angles modulo 180°. Let G' be the orthogonal projection of the point D on the line BE. Then, < BG'D = 90°. On the other hand, < BFD = 90°, since $DF\perp AB$. Thus, the points G' and F lie on the circle with diameter BD; hence, < BG'F = < BDF. But since $DF\perp AB$, we have < (AB; DF) = 90°, and since $AH\perp BC$, we have < (AH; BC) = 90°. Thus, < HG'F = < BG'F = < BDF = < (BD; DF) = < (BD; AB) + < (AB; DF) = < (BD; AB) + 90° = < (BD; AB) + < (AH; BC) = < (BD; AB) + < (AH; BD) = < (AH; AB) = < HAF. Hence, the point G' lies on the circumcircle of triangle AFH. In other words, the point G' is the point of intersection of the circumcircle of triangle AFH with the line BE different from the point H. Thus, the point G' coincides with the point G. Hence, < BG'D = 90° becomes < BGD = 90°, and we are done. darij

I have a nice solution to prove that BGD=90 and it is dedicated to Darij (my figure is different from yetti I have AC>AB) Let the altitude from A intersects BC at K . Then from the cyclic FKDA we have that $BF\cdot BA=BK\cdot BD$ . But from the power point theorem we have that $BH\cdot BG=BF\cdot BA$ so $BH\cdot BG=BK\cdot BD$ so the triangles BHK and BGD are similar and we are done PS I hope you like it