<p>In triangle \(ABC\), \(AE\) and \(BF\) are perpendicular to \(BC\) and \(AC\), respectively. \(D\) is a point on \(AB\) such that&nbsp;\(COD\) is a straight line.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/public/question-images/pasted/1775801018491-bujryzu9tvc.png" style="max-width: min(270px, 100%); height: auto; display: block;" alt="Diagram showing triangle ABC with altitudes AE and BF intersecting at O, and point D on AB." class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">Prove the concurrency (intersecting at one point) of the \(3\) altitudes (heights) of a triangle by proving that&nbsp;\(CD\) is perpendicular to \(AB\).</span></div></li></ol>
<p>Hint: Draw lines that are parallel to \(BC\), \(AC\) and \(AB\) through points \(A\), \(B\) and \(C\), respectively, then use the fact that the \(3\) perpendicular bisectors of the sides of a triangle intersect at one point to prove that&nbsp;\(CD\) is perpendicular to \(AB\).</p>
<ol class="i-list" start="3"><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">Hence prove that there exists a circle with a diameter \(OB\) that passes through the \(4\) points \(O\), \(D\), \(B\) and \(E\).</span></div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">From the diagram above, state another group of \(4\) points that lie on the same circle and state one diameter of the circle.</span></div></li></ol>
<p>Note: The point \(O\) is known as the <strong>orthocentre</strong> of triangle \(ABC\) and there are totally \(6\) groups of \(4\) points in the above diagram that lie on the same circle.</p>

In triangle \(ABC\), \(AE\) and \(BF\) are perpendicular to \(BC\) and \(AC\), respectively. \(D\) is a point on \(AB\) such that \(COD\) is a straight line.

Prove the concurrency (intersecting at one point) of the \(3\) altitudes (heights) of a triangle by proving that \(CD\) is perpendicular to \(AB\).

Hint: Draw lines that are parallel to \(BC\), \(AC\) and \(AB\) through points \(A\), \(B\) and \(C\), respectively, then use the fact that the \(3\) perpendicular bisectors of the sides of a triangle intersect at one point to prove that \(CD\) is perpendicular to \(AB\).

Hence prove that there exists a circle with a diameter \(OB\) that passes through the \(4\) points \(O\), \(D\), \(B\) and \(E\).

From the diagram above, state another group of \(4\) points that lie on the same circle and state one diameter of the circle.

Note: The point \(O\) is known as the **orthocentre** of triangle \(ABC\) and there are totally \(6\) groups of \(4\) points in the above diagram that lie on the same circle.

Concurrency of the Altitudes (Heights) of Triangles and the Orthocentre of Triangles

Concurrency of the Altitudes (Heights) of Triangles and the Orthocentre of Triangles