Tag Archives: circle geometry

May 24, 2025 · 10:02 am

Circle Geometry Problem

In the diagram, points $A, B, C, D$ and $Q$ lie on a circle centre $O$ , radius $6$ cm and diameter $BQ, \angle{ABQ}=50^\circ, AB$ is parallel to $DO$ and point $P$ lies on diameter $BQ$ such that $OP=DP=4$ cm.

(a) Find $\angle{BCD}$

(b) Determine the length of $PC$ .

$\angle{BOD}=180^\circ-50^\circ=130^\circ$ (Co-interior angles in parallel lines are supplementary.)

$\angle{BCD}=\frac{1}{2}\times 130=65^\circ$ (Angles subtended by the same arc. The angle at the centre is twice the angle at the circumference.)

$\angle{BCD}=65^\circ.$

Let $PC=x$

From the intersecting chord theorem

$\begin{equation*}4\times x=2\times 10\end{equation}$

$\begin{equation*}4x=20\end{equation}$

$\begin{equation*}x=5\end{equation}$

$PC=5cm$

A chord $AB$ of a circle $O$ is extended to $C$ . The straight line bisecting $\angle{OAB}$ meets the circle at $E$ . Let $\angle{BAE}=x$ . Prove that $EB$ bisects $\angle{OBC}$ .

$\angle{BAO}=2x$ ( $AE$ bisects $\angle {BAO}$ )

$\Delta AOB$ is isosceles ( $AO=B0$ radii of the circle)

$\angle{ABO}=2x$ (Equal angles in isosceles triangle)

Therefore $\angle {AOB}=180^\circ-4x$ (angle sum of a triangle)

$\angle {BEA}=90^\circ-2x$ (angle at the circumference is half angle at the centre)

$\angle{ABE}=180^\circ-x-(90^\circ-2x)=90^\circ+x$ (angle sum of a triangle)

$\angle{CBE}=180^\circ-(90^\circ+x)=90^\circ-x$ (angles on a straight line)

$\angle{OBE}=90^\circ+x-2x=90^\circ-x$

$\angle{OBE}=\angle{CBE}$

Hence, $BE$ bisects $\angle{OBC}$

Intersecting Secant Theorem

$CD$ is a tangent to the circle.

Prove $c^2=a(a+b)$

I am going to add two chords to the circle

$\angle{BDC}=\angle{CAD}$ (angles in alternate segments are congruent)

$\angle{BCD}=\angle{DCA}$ (shared angle)

$\therefore \Delta BDC\cong \Delta{DAC}$ (AA)

Hence

$\frac{DC}{AC}=\frac{BC}{DC}$ (Corresponding sides in similar triangles)

$\frac{c}{a+b}=\frac{a}{c}$

$\therefore c^2=a(a+b)$

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Filed under Circle Theorems, Geometry, Year 11 Specialist Mathematics

Tagged as circle geometry, Intersecting Secants Theorem

September 7, 2024 · 9:31 am

Geometry Problem

This problem is from Geometry Snacks by Ed Southall and Vincent Pantaloni – it’s a great book.

Two squares are constructed such that three vertices are collinear as shown. Find the value of the marked angle.

I started by marking in the right angles. And I added the diagonal of the larger square (pink line).

Because there are right angles at $O$ and $P$ , we know there is a circle, which has the diagonal of the square as its diameter (see second image below).