Area of triangle formed by circumcircle of triangle an altitude of opposite vertex

Question

Consider triangle ABC with AB=3, AC=5, BC=7, where the altitude of A intersects the circumcircle of triangle ABC at X, which is different from A. What would be the area of triangle XBC?

I don't think coordinates would be a good idea, but I honestly do not have any clue to start? Perhaps I am unfamiliar with a geometry theorem needed for this question?

Thanks for any advice you may have!

Answer 1

Hints-(1). Apply cosine law twice to find $\angle ABC$ and aslo $\angle ACB$.

(2). Let AX cut BC at P. (From $\triangle ABP$) find PB and (from $\triangle APC$) find PC.

(3). Since ABXC is cyclic, $\angle ABC = \angle PXC$.

(4). Find XP.

Required result can then be found.

Answer 2

Let $|AB|=3=c,\ |AC|=5=b,\ |BC|=7=a$.

Area of $\triangle ABC$ is

\begin{align} S_{ABC}&=\tfrac14\,\sqrt{4a^2b^2-(a^2+b^2-c^2)^2} =\tfrac{15}4\,\sqrt3 ,\\ |AH|&=\frac{2S_{ABC}}a =\tfrac{15}{14}\,\sqrt3 ,\\ |BH|&= \sqrt{c^2-|AH|^2} =\tfrac{33}{14} ,\\ |CH|&=|BC|-|BH|= =\tfrac{65}{14} . \end{align}

By the power of the point $H$ wrt the circumcircle, \begin{align} |BH|\cdot|CH| &=|AH|\cdot|HX| ,\\ |HX| &= \frac{|BH|\cdot|CH|}{|AH|} = \tfrac{143}{42}\,\sqrt3 . \end{align}

$|HX|$ is the altitude of $\triangle XBC$,

so the answer is \begin{align} S_{XBC}&=\tfrac12\,|BC|\cdot|HX| =\tfrac{143}{12}\,\sqrt3 . \end{align}

Answer 3

In general,

$${K_{XBC}} = K_{ABC}\cot B\cot C$$

which leads to

$$K_{XBC}=\frac{(a^2+b^2-c^2)(a^2+c^2-b^2)} {4\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}}$$

where $a,\>b$ and $c$ are the side lengths of the triangle ABC.