1. circle geometry help 2

2. Hello andy69

The question is shown in the diagram I have attached.

Proof

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

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

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

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

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

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

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

But $\displaystyle BQ = XQ$ (proven)

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

$\displaystyle \therefore BQ = QC$

Hello andy69

The question is shown in the diagram I have attached.

Proof

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

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

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

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

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

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

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

But $\displaystyle BQ = XQ$ (proven)

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

$\displaystyle \therefore BQ = QC$