$\begin{align}&\text{Construction - Draw OX Perpendicular from O to AB}\\[5pt] &AB \perp OX , \text { }\text{X is the midpoint of AB}\\[5pt] AB &=12cm\\[5pt] XB &=\frac{1}{2}\times 12=6cm\\[5pt] &\text{Applying Pythagoras' theorem to OXB triangle}\\[5pt] OB^2 &=OX^2+XB^2\\[5pt] &=8^2+6^2\\[5pt] &=64+36\\[5pt] &=100\\[5pt] OB &=\sqrt{100}\\[5pt] &\underline{\underline{OB=10cm}} \end{align}$
$\begin{align}O\hat XB &=O\hat YB=90^0\text { }\text{Since the line joining the center of the circle to the midpoint of a chord is perpendicular to the chord.}\\[5pt] AB&=6cm\\[5pt] XB&=3cm\text{(X is the midpoint of AB)}\\[5pt] BC&=8cm\\[5pt] BY&=4cm\text{(Y is the midpoint of BC)}\\[5pt] &\text{Applying Pythagoras' theorem to OBY triangle}\\[5pt] OB^2 &=OY^2+YB^2\\[5pt] 5^2&=OY^2+4^2\\[5pt] OY^2&=5^2-4^2\\[5pt] &=25-16\\[5pt] &=9\\[5pt] OY &=\sqrt{9}\\[5pt] &\underline{\underline{OY=3cm}} \end{align}$ $\begin{align}&\text{Applying Pythagoras' theorem to OXB triangle}\\[5pt] OB^2 &=OX^2+XB^2\\[5pt] 5^2&=OX^2+3^2\\[5pt] OX^2&=5^2-3^2\\[5pt] &=25-9\\[5pt] &=16\\[5pt] OX &=\sqrt{16}\\[5pt] &\underline{\underline{OX=4cm}}\\[5pt] &\text{Perimeter of the OXBY}\\[5pt] &=OY+YB+BX+XO\\[5pt] &=3+4+3+4\\[5pt] &\underline{\underline{14cm}} \end{align}$
$\begin{align}&AB \perp OX , \text { }\text{X is the midpoint of AB}\\[5pt] AB &=8cm\\[5pt] AX &=4cm\text{(X is the midpoint of AB)}\\[5pt] &\text{Applying Pythagoras' theorem to OXA triangle}\\[5pt] OA^2 &=OX^2+XA^2\\[5pt] r^2&=(r-3)^2+4^2\\[5pt] r^2&=r^2-6r+9+16\\[5pt] 6r&=25\\[5pt] r &=\frac{25}{6}\\[5pt] &\underline{\underline{r=4\frac{1}{6}cm}} \end{align}$
$\begin{align}A\hat XC &=B\hat XC=90^0\text { }\text{Since the line joining the center of the circle to the midpoint of a chord is perpendicular to the chord.}\\[5pt] &AXC \triangle \text { }\text{and} \text { }BXC \triangle\\[5pt] AX &=XB\text{(Data)}\\[5pt] A\hat XC&=B\hat XC\text{ } (90^0)\\[5pt] CX &=CX\text{(Common)}\\[5pt] &AXC \triangle \equiv BXC \triangle\text{(SAS)}\\[5pt] &\underline{\underline{AC=BC\text { }\text{(Corresponding elements of congruent triangles are equal)}}} \end{align}$
$\begin{align}&OXQ \triangle \text { }\text{and} \text { }OYR \triangle\\[5pt] X\hat OQ&=Y\hat OR\text{(Opposite angles )} \\[5pt] O\hat XQ&=O\hat YR\text{ } (90^0)\\[5pt] OQ &=OR\text{(Radius)}\\[5pt] &OXQ \triangle \equiv OYR \triangle\text{(AAS)}\\[5pt] XQ&=YR\text { }\text{(Corresponding elements of congruent triangles are equal)}\\[5pt] 2XQ&=2YR-->(1)\\[5pt] &PQ \perp OX , \text { }\text{X is the midpoint of PQ}\\[5pt] 2XQ&=PQ-->(2)\\[5pt] &PR \perp OY , \text { }\text{Y is the midpoint of PR}\\[5pt] 2YR&=PR-->(3)\\[5pt] &(2) , (3)\text { }\text{Substitute in (1)}\\[5pt] &\underline{\underline{PQ=PR}} \end{align}$
$\begin{align}\text{Construction} &-\text{ Draw OX Perpendicular from O to AB}\\[5pt] &-\text{Draw OY Perpendicular from O to CD}\\[5pt] &AB \perp OX , \text { }\text{X is the midpoint of AB}\\[5pt] &CD \perp OY , \text { }\text{Y is the midpoint of CD}\\[5pt] AB &=24cm\\[5pt] XB &=\frac{1}{2}\times 24=12cm\\[5pt] &\text{Applying Pythagoras' theorem to OXB triangle}\\[5pt] OB^2 &=OX^2+XB^2\\[5pt] &=5^2+12^2\\[5pt] &=25+144\\[5pt] &=169\\[5pt] OB &=\sqrt{169}\\[5pt] &\underline{\underline{OD=OB=13cm}}\\[5pt] &\text{Applying Pythagoras' theorem to OYD triangle}\\[5pt] OD^2 &=OY^2+YD^2\\[5pt] YD^2&=13^2-12^2\\[5pt] &=169-144\\[5pt] &=25\\[5pt] YD &=\sqrt{25}\\[5pt] &\underline{\underline{YD=5cm}}\\[5pt] CD&=2YD\text { }\text{(Y is the midpoint of CD)}\\[5pt] &=2\times 5\\[5pt] &\underline{\underline{CD=10cm}} \end{align}$
$\begin{align}&AB \perp OX , \text { }\text{X is the midpoint of AB}\\[5pt] AX &=\frac{1}{2}\times AB\\[5pt] &AC \perp OY , \text { }\text{Y is the midpoint of AC}\\[5pt] AY &=\frac{1}{2}\times AC\\[5pt] AB &=AC\text{(ABC is an equivalent triangle)}\\[5pt] \frac{1}{2}\times AB&=\frac{1}{2}\times AC\\[5pt] AX &=AY\\[5pt] &AOX \triangle \text { }\text{and} \text { }AOY \triangle\\[5pt] A\hat XO&=A\hat YO\text{ } (90^0)\\[5pt] AX &=AY\text{(Proved)}\\[5pt] AO &=AO\text{(Common)}\\[5pt] &AOX \triangle \equiv AOY \triangle\text{(RHS)}\\[5pt] OX&=OY--(1)\text { }\text{(Corresponding elements of congruent triangles are equal)}\\[5pt] &YOC \triangle \equiv ZOC \triangle \\[5pt] OY&=OZ--(2)\text { }\text{(Corresponding elements of congruent triangles are equal)}\\[5pt] &\text{Using (1) and (2)}\\[5pt] &\underline{\underline{OX=OY=OZ}} \end{align}$
$\begin{align}&PQ \perp OX , \text { }\text{X is the midpoint of PQ}\\[5pt] PX &=\frac{1}{2}\times PQ\\[5pt] &\text{Applying Pythagoras' theorem to PXO triangle}\\[5pt] PO^2 &=PX^2+OX^2\\[5pt] &=(\frac{1}{2}PQ)^2+OX^2\\[5pt] &=\frac{1}{4}PQ^2+OX^2 -->(1)\\[5pt] &RS\perp OY , \text { }\text{Y is the midpoint of RS}\\[5pt] RY&=(\frac{1}{2})RS\\[5pt] &\text{Applying Pythagoras' theorem to RYO triangle}\\[5pt] RO^2 &=RY^2+OY^2\\[5pt] RO^2&=(\frac{1}{2}RS)^2+OY^2\\[5pt] &=\frac{1}{4}RS^2+OY^2 -->(2)\\[5pt] PO&=RO\text{(Radius of the circle)}\\[5pt] PO^2&=RO^2\\[5pt] &(1)=(2)\\[5pt] \frac{1}{4}PQ^2+OX^2&= \frac{1}{4}RS^2+OY^2\\[5pt] PQ^2+4OX^2&=RS^2+4OY^2\\[5pt] &\underline{\underline{PQ^2-RS^2=4OY^2-4OX^2}} \end{align}$

By studying this lesson you will be able to

Solve problems by using the theorem that the straight line joining the midpoint of a chord of a circle to the centre is perpendicular to the chord.