**Topic**

For acute triangle *ABC *have different double sides. Call *H*, *O* are the orthocenter and circumcenter of the triangle, respectively, *EASY* is the point of symmetry with *H* via *O*. Prove \(\overrightarrow {HA} + \overrightarrow {HB} + \overrightarrow {HC} = \overrightarrow {HD} \)

**Solution method – See details**

Step 1: Get *E* symmetrical with *A* via *O*

Step 2: Prove the quadrilaterals *ADEH*, *BHCE* is a parallelogram

Step 3: Apply parallelogram rule to prove \(\overrightarrow {HA} + \overrightarrow {HB} + \overrightarrow {HC} = \overrightarrow {HD} \)

**Detailed explanation**

Call *E* is the point of symmetry with *A* via *O* . Then AE is the diameter of the circumcircle .*ABC*

Quadrilateral *ADEH *yes *O *is the midpoint *HD* and *AE* should be a parallelogram

\( \Rightarrow \overrightarrow {HA} + \overrightarrow {HE} = \overrightarrow {HD} \)(1)

Again: \(\widehat {ACE}\) is an inscribed angle intercepting the semicircle, so \(\widehat {ACE} = {90^0}\)\( \Rightarrow EC \bot AC\), where \( BH \bot AC\)

\( \Rightarrow EC//BH\)

Similar proof we have \(BE//HC\)

Quadrilateral BHCE has \(EC//BH\), \(BE//HC\) so it is a parallelogram

\( \Rightarrow \overrightarrow {HB} + \overrightarrow {HC} = \overrightarrow {HE} \)(2)

From (1) and (2) deduce \(\overrightarrow {HA} + \overrightarrow {HB} + \overrightarrow {HC} = \overrightarrow {HA} + \overrightarrow {HE} = \overrightarrow {HD} \) ( PCM)