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Belarus — Selection and Training Session

Geometry Difficulty 6.1 National Olympiad Prove it Belarus

The external angle bisector of the angle $A$ of an acute-angled triangle $ABC$ meets the circumcircle of $\triangle ABC$ at point $T$ . The perpendicular from the orthocenter $H$ of $\triangle ABC$ to the line $TA$ meets the line $BC$ at point $P$ . The line $TP$ meets the circumcircle of $\triangle ABC$ at point $D$ .

Prove that $AB^2 + DC^2 = AC^2 + BD^2$ .

Solution

Let $l_a$ be the bisectrix of $\angle A$ . Then $l_a \perp AT$ and $PH \parallel l_a$ (as, by condition, $PH \perp AT$ ). Further,
$\angle (BH, HP) = \angle (BH, AC) + \angle (AC, HP) = 90^\circ + \angle (AC, l_a) = 90^\circ + \angle (l_a, AB) = 90^\circ + \angle (HP, AB) = \angle (HP, AB) + \angle (AB, CH) = \angle (HP, CH).$
Hence, $HP$ is the bisectrix of $\angle BHC$ .

Let $AH$ meet the circumcircle of $\triangle ABC$ at $D_1 \neq A$ . We have $\angle D_1BC = 90^\circ - \angle C = \angle HBC$ and, similarly, $\angle D_1CB = \angle HCB$ . Then the triangles $BHC$ and $BD_1C$ are equal, Hence, $D_1P$ is the bisectrix of $\angle BD_1C$ . Therefore, $D_1P$ intersects the circumcircle at the middle of the arc $BAC$ , i.e. at the point $T$ . So, $D_1 = D$ and $AD \perp BC$ . Thus the statement follows from the well-known

Lemma. The diagonals of the quadrilateral $ABCD$ are perpendicular if and only if $AB^2 + CD^2 = AC^2 + BD^2$ .

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