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Solving Lesson 46 Page 92 SBT Math 10 – Kite>


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)



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