PBA and PDC are two secants. AD is the diamet | vedclass

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(B) $1$. Since $AD$ is the diameter,$\angle ABD = 90^{\circ}$ (angle in a semicircle).
$2$. In $	riangle PBD$,the sum of angles is $180^{\circ}$. Th

Vedclass · Accepted Answer

Quantity $I < $ Quantity $II$

Vedclass · Answer

(B) $1$. Since $AD$ is the diameter,$\angle ABD = 90^{\circ}$ (angle in a semicircle).
$2$. In $	riangle PBD$,the sum of angles is $180^{\circ}$. Thus,$\angle P + \angle PBD + \angle BDP = 180^{\circ}$. Since $\angle PBD = 90^{\circ}$ (as $P, B, A$ are collinear and $\angle ABD = 90^{\circ}$),we have $20^{\circ} + 90^{\circ} + \angle BDP = 180^{\circ}$,so $\angle BDP = 70^{\circ}$.
$3$. In $	riangle ABD$,$\angle ADB = 180^{\circ} - \angle DAB - \angle ABD = 180^{\circ} - 40^{\circ} - 90^{\circ} = 50^{\circ}$. This is Quantity $II$.
$4$. $\angle BCD = \angle DAB = 40^{\circ}$ (angles subtended by the same arc $BD$ in the same segment).
$5$. In $	riangle PBC$,the sum of angles is $180^{\circ}$. $\angle P + \angle PBC + \angle BCP = 180^{\circ}$. Note that $\angle PBC = 180^{\circ} - \angle ABD = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
$6$. Thus,$20^{\circ} + 90^{\circ} + \angle BCP = 180^{\circ} \Rightarrow \angle BCP = 70^{\circ}$.
$7$. Since $P, C, D$ are collinear,$\angle BCD + \angle BCP = 180^{\circ}$. $40^{\circ} + 70^{\circ} = 110^{\circ} 
eq 180^{\circ}$. Re-evaluating: $\angle DBC = 180^{\circ} - \angle BCD - \angle BDC = 180^{\circ} - 40^{\circ} - 70^{\circ} = 70^{\circ}$.
$8$. Actually,$\angle DBC = 30^{\circ}$ as calculated from $	riangle BCD$ where $\angle BCD = 40^{\circ}$ and $\angle BDC = 180^{\circ} - 70^{\circ} = 110^{\circ}$.
$9$. Since Quantity $I = 30^{\circ}$ and Quantity $II = 50^{\circ}$,Quantity $I < $ Quantity $II$.

$PBA$ and $PDC$ are two secants. $AD$ is the diameter of the circle with center at $O$. $\angle A = 40^{\circ}$,$\angle P = 20^{\circ}$. Find the relationship between Quantity $I$ $(\angle DBC)$ and Quantity $II$ $(\angle ADB)$.

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