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<p>The diagram shows a circle, centre $O$. Points $A$, $B$, $D$ and $E$ lie on the circle. <br>The line $AD$ is a diameter of the circle and $DCP$ is the tangent of the circle at $D$. <br>The lines $EB$ and $AD$ intersect at point $Q$. <br>The chord $AB$ is produced to meet the tangent $DCP$ at $P$. <br>It is given that&nbsp;$\text{angle}\hspace{0.5em} CBP = \text{angle}\hspace{0.5em} DEB$.</p>
<ol class="abc-list"><li data-list-item-id="item-1"><div class="list-item-paragraph">Show that&nbsp;$\text{angle}\hspace{0.5em} BCD = 90^\circ$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-2"><div class="list-item-paragraph">Show that&nbsp;$DP^2 = BP^2 + DA \times BC$.<span class="inline-marks">[4]</span></div></li></ol>

The diagram shows a circle, centre $O$. Points $A$, $B$, $D$ and $E$ lie on the circle. 
The line $AD$ is a diameter of the circle and $DCP$ is the tangent of the circle at $D$. 
The lines $EB$ and $AD$ intersect at point $Q$. 
The chord $AB$ is produced to meet the tangent $DCP$ at $P$. 
It is given that $\text{angle } CBP = \text{angle } DEB$.

Show that $\text{angle } BCD = 90^\circ$.

2024 CGSS P1 Q13

2024 CGSS P1 Q13