# Thread: circle geometry help 2

1. ## circle geometry help 2

2. Hello andy69

The question is shown in the diagram I have attached.

Proof

(i) $\angle QXB + \angle BXA + \angle AXP = 180^o$ (angles on a straight line)

$\angle BXA = 90^o$ (given)

$\therefore \angle QXB + \angle AXP = 90^o$

But $\angle PAX + \angle AXP = 90^o$ (angle sum of $\triangle APX$)

$\therefore \angle QXB = \angle PAX$
$=\angle QBX$ (angles in same segment)
(ii) $BQ = XQ$ (isosceles $\triangle BQX$, from part (i))

$\angle BXC = 90^o$ (given)

$\therefore X$ lies on a circle with diameter BC (angle in semicircle)

But $BQ = XQ$ (proven)

$\therefore Q$ is the centre of the circle through $B, \;X$ and $C$

$\therefore BQ = QC$

Hello andy69

The question is shown in the diagram I have attached.

Proof

(i) $\angle QXB + \angle BXA + \angle AXP = 180^o$ (angles on a straight line)

$\angle BXA = 90^o$ (given)

$\therefore \angle QXB + \angle AXP = 90^o$

But $\angle PAX + \angle AXP = 90^o$ (angle sum of $\triangle APX$)

$\therefore \angle QXB = \angle PAX$
$=\angle QBX$ (angles in same segment)
(ii) $BQ = XQ$ (isosceles $\triangle BQX$, from part (i))

$\angle BXC = 90^o$ (given)

$\therefore X$ lies on a circle with diameter BC (angle in semicircle)

But $BQ = XQ$ (proven)

$\therefore Q$ is the centre of the circle through $B, \;X$ and $C$

$\therefore BQ = QC$