Sec 4 A Math: Proofs in Plane Geometry, practice questions & worked solutions

$$\begin{aligned} \text{angle } TPR &= \text{angle } TRP (TP = TR,\ \triangle TPR \text{ is an isosceles})\\ \text{angle } TPQ &= \text{angle } TSP (\text{alternate segment thm.}) \end{aligned}$$ $$\begin{aligned} \text{angle } QPR &= \text{angle } TPR - \text{angle } TPQ \end{aligned}$$ $$\begin{aligned} \text{angle } TRP &= \text{angle } TSP + \text{angle } RPS (\text{exterior angle of } \triangle) \end{aligned}$$ $$\begin{aligned} \text{angle } RPS &= \text{angle } TRP - \text{angle } TSP\\ &= \text{angle } TPR - \text{angle } TPQ\\ &= \text{angle } QPR \end{aligned}$$ $\therefore$ line $PR$ bisects the angle $SPQ$. (proven)

By midpoint theorem, $DE$ is parallel to $BC$.

Thus, $\angle ADE = \angle ABC$. (corresponding angles)

By tangent-chord theorem, $$\angle ABC = \angle ECT$$

Since $\angle ADE = \angle ECT$, they are angles in same segment, hence, points $A$, $D$, $C$ and $T$ lie on a circle.

(a)

Method 1: $$\begin{aligned} \angle BAC &= \angle BCQ \text{(Alt.\ Seg.\ Thm.)}\\ \angle BOC &= 2(\angle BAC) (\angle \text{ at centre}=2\angle\text{s at circumference})\\ &= 2\angle BCQ \end{aligned}$$

Method 2: Let $\angle BCQ=x$. $$\begin{aligned} \angle OCB &= 90^\circ-x \text{(rad.\ } \perp \text{ tan)}\\ \angle BOC &= 180^\circ-2(90^\circ-x) \text{(isosceles }\triangle OBC)\\ &= 180^\circ-180^\circ+2x\\ &= 2x\\ &= 2\angle BCQ \end{aligned}$$

(b)

(a) $$\begin{aligned} \angle OPC &= \angle QPB \text{(common pt }P)\\ \angle PCO &= \angle PBQ = 90^\circ \text{(rad.\ } \perp \text{ tan)} \end{aligned}$$ Since 2 pairs of corresponding angles are equal, triangle $POC$ is similar to triangle $PQB$.

(b)

(b) Since triangle $POC$ is similar to triangle $PQB$, $$\frac{PO}{PQ}=\frac{PC}{PB}.$$ Since $C$ is the midpoint of $PQ$, $\dfrac{PC}{PQ}=\dfrac{1}{2}\;\Rightarrow\;PC=\dfrac{1}{2}PQ$. $$\begin{aligned} \frac{PO}{PQ} &= \frac{\frac{1}{2}PQ}{PB}\\ PO\times PB &= \frac{1}{2}PQ\times PQ\\ &= \frac{1}{2}PQ^2 \end{aligned}$$

(i)

$$\begin{aligned} \angle ATD &= \angle XAB (\text{corr angles, } AB /\!/ TD)\\ &= \angle ACB (\text{alternate segment theorem}) \end{aligned}$$

(ii)

Since $\angle ATD = \angle ACB$, they fulfil the property of angles in the same segment. $$\therefore \text{A circle can be drawn passing through the points } T, A, D \text{ and } C.$$

(iii)

$$\begin{aligned} \angle ATE &= \angle DCE [\text{from part (a)}]\\ \angle TEA &= \angle CED [\text{vert opp } \angle]\\ \therefore \triangle ATE &\text{ is similar to } \triangle DCE (\text{AA similarity test}) \end{aligned}$$ $$\begin{aligned} \frac{TE}{CE} &= \frac{EA}{ED} (\text{Ratio of corr sides of similar triangles are equal})\\ CE \times AE &= DE \times TE (\text{proven}) \end{aligned}$$

(a)

$$\begin{aligned} \angle RSU &= \angle RUS (RS = RU)\\ &= \angle URQ (\text{alt seg thm})\\ \angle URQ &= \angle RUQ\\ \therefore \triangle RQU &\text{ is isosceles} \end{aligned}$$

(b)

$$\begin{aligned} \angle RQS &= 2\angle QRU (\text{ext } \angle \text{ of a triangle}, \triangle RQU \text{ is isosceles})\\ \angle RQS &= \angle SPT (\text{angles in the same seg})\\ \angle QRU &= \angle QPT (\text{alt seg theorem})\\ \therefore \angle SPT &= 2\angle QPT \end{aligned}$$

(a)

Let $x = \text{Angle } QPA$. $$\begin{aligned} \text{angle } APB &= x (AP \text{ bisects angle } QPB)\\ \text{angle } ABP &= x (\text{alt.\ seg thm}) \end{aligned}$$ Since angle $APB =$ angle $ABP = x$,

Triangle $ABP$ is isosceles.

(b)

(i) $$\begin{aligned} \text{angle } ACP &= x (\angle \text{ in the same seg})\\ \text{angle } ACP &= \text{angle } APD = x\\ \text{angle } CAP &= \text{angle } PAD (\text{common } \angle) \end{aligned}$$ Triangle $ACP$ and Triangle $APD$  (AA Similarity Test)

(ii) $$\begin{aligned} \frac{AD}{AP} &= \frac{AP}{AC}\\ AP^2 &= AC \times AD (\because AP = AB)\\ AB^2 &= AC \times AD \end{aligned}$$

(a)

$$\begin{aligned} \angle YXC &= \angle CBX \text{(tangent chord theorem)}\\ \angle YXC &= \angle CXF (XC \text{ is an angle bisector of } YXB)\\ \angle CXF &= \angle CBX \end{aligned}$$ $\triangle BCX$ is an isoceles triangle, $CX=CB$.

(b)

$$\begin{aligned} \angle AXC &= 90^\circ \text{(right angle in a semicircle)}\\ \angle BFC &= 90^\circ\\ & (AC \text{ is a chord bisector of } BX \text{ because } \triangle BCX \text{ is an isoc triangle (a)})\\ \angle AXC &= \angle BFC\\ \angle XBC &= \angle XAC = \angle FBC \text{(angles in the same segment)} \end{aligned}$$ $AA$ similarity, triangle $AXC$ is similar to triangle $BFC$.

(c)

triangle $AXC$ is similar to triangle $BFC$ $$\begin{aligned} \frac{AC}{BC} &= \frac{XC}{FC}\\ AC\times FC &= XC\times BC\\ XC &= BC \text{(from part (a))}\\ AC\times FC &= BC^2 \end{aligned}$$

(a)

$$\begin{aligned} \angle PQC &= \angle PSQ \text{(Alternate segment theorem)}\\ \angle PQC &= \angle PBC \text{(angles in same segment)} \end{aligned}$$ Hence $\angle PBC=\angle PSQ$. By alternate angles of parallel lines, $BC$ is parallel to $QS$.

(b)

$$\begin{aligned} \angle PQC &= \angle PSQ \text{(Alternate segment theorem)}\\ \angle PMQ &= \angle QMS \text{(Common angle)} \end{aligned}$$ $\triangle PQM$ is similar to $\triangle QSM$ (AA Similarity Test).

(c)

Since $\triangle PQM$ is similar to $\triangle QSM$, $\dfrac{QM}{SM}=\dfrac{PM}{QM}$. $$(QM)^2 = PM\times SM$$

(a)

$$\begin{aligned} \angle ACB &= 90^\circ \text{(Angles in semicircle)}\\ \angle ABF &= 90^\circ \text{(Radius perpendicular to tangent)}\\ \angle ACB &= \angle ABF\\ \angle CAB &= \angle BAF \text{(common angle)} \end{aligned}$$ Triangle $ABC$ and triangle $AFB$ are similar (AA similarity).

(b)

Triangle $ABC$ and triangle $AFB$ are similar. $$\begin{aligned} \angle ABC &= \angle AFB\\ \angle CFE &= \angle AFB \text{(same angle)}\\ \angle ABC &= \angle DCA \text{(tangent chord theorem)}\\ \angle DCA &= \angle FCE \text{(vertically opposite angle)}\\ \angle FCE &= \angle ABC = \angle AFB \end{aligned}$$ Therefore $\angle CFE=\angle FCE$, triangle $CFE$ is isosceles, $EC=EF$.

Alternative solution: $EB=EC$ (common tangent). Triangle $ECB$ is isosceles. Let $\angle EBC=x$. $$\begin{aligned} \angle ECB &= x\\ \angle FCE &= 180-90-x \text{(Adjacent angles in a straight line)} = 90-x\\ \angle CAB &= \angle ECB \text{(tangent chord theorem)} = x\\ \angle CFE &= 180-90-x \text{(angle sum of triangle)} = 90-x = \angle FCE \end{aligned}$$ Therefore triangle $CFE$ is isosceles, $EC=EF$.

(a)

In $\triangle ABE$ and $\triangle CBE$,

$AB = BC$ (given)

$AE = EC$ ($E$ is the midpoint of $AC$)

$BE$ is a common height

$\triangle ABE$ is congruent to $\triangle CBE$. (SSS)

(b)

Since $\triangle ABE$ is congruent to $\triangle CBE$, $\angle BEA = \angle BEC$. (corresponding angles of congruent triangles) $$\begin{aligned} \angle BEA + \angle BEC &= 180^\circ (\text{angles on a straight line})\\ \angle BEA + \angle BEA &= 180^\circ\\ \angle BEA &= 90^\circ \end{aligned}$$

(c)

Since $D$ and $E$ are midpoints of $AB$ and $AC$ respectively, by midpoint theorem, $DE \parallel BC$ and $DE = \dfrac{1}{2}BC$.

In $\triangle ADE$ and $\triangle ABC$,

$\angle DAE = \angle BAC$ (common angle)

$\angle ADE = \angle ABC$ (corresponding angle, $DE \parallel BC$)

$\triangle ADE$ is similar to $\triangle ABC$. (AA similarity test)

(d)

In $\triangle DFE$ and $\triangle CFB$,

$\angle EFD = \angle BFC$ (vertically opposite angle)

$\angle FDE = \angle FCB$ (alternate angle, $DE \parallel BC$)

$\triangle DFE$ is similar to $\triangle CFB$.

By midpoint theorem, $DE = \dfrac{1}{2}BC$. $$\begin{aligned} \frac{DF}{FC} &= \frac{DE}{BC}\\ &= \frac{1}{2} (\text{proven}) \end{aligned}$$