Question

In the figure given below,$C$ and $D$ are centres of two intersecting circles.The line $APQB$ is perpendicular to the line of centres $CD$

$AP=QB$ and $AQ=BP$

• A. True
• B. False

Answer 1

The correct option is A True

Join $AC$ and $BC,DP$ and $DQ$

In right angled $△ACM$ and $△BCM$

hypotenuse$AC=BC$ (radii of same circle)

Side $CM=CM$(common)

$\therefore △ACM\cong △BCM$ by R.H.S axiom of congruency

$\therefore AM=BM$ .......$\left(1\right)$

Again in right angled $△PDM$ and $△QDM$

Hyp$PD=QD$ (radii of the same circle)

Side $DM=DM$ (common)

$\therefore △PDM\cong △QDM$ by R.H.S axiom of congruency

$\therefore PM=QM$ ......$\left(2\right)$

Subtracting $\left(2\right)$ from $\left(1\right)$ we get

$AM-PM=BM-QM$

$\Rightarrow AP=QB$

Adding $PQ$ both sides,

$AP+PQ=PQ+AB$

$\Rightarrow AQ=PB$