Question: In the figure, $AB \parallel DE$. Prove that $\triangle ABC$ is similar to $\triangle EDC$.

Step 1. $\angle ACB = \angle ECD$ (vertically opposite angles).

Step 2. $\angle BAC = \angle DEC$ (alternate angles, $AB \parallel DE$).

Step 3. Two pairs of equal angles → by AA, $\triangle ABC \sim \triangle EDC$.

Tip: AA (two equal angles) is enough for similarity. Once you have AA, the third pair is automatic ($\angle ABC = \angle EDC$).

Question: $ABCD$ is a parallelogram whose diagonals $AC$ and $BD$ meet at $E$. Prove that $\triangle ABE \equiv \triangle CDE$, and hence that the diagonals bisect each other.

Step 1. Match a pair of sides:

$$AB = CD \;\text{(opposite sides of a parallelogram)}$$

Step 2. First pair of angles — use the parallel pair $AB \parallel DC$ with transversal $AC$:

$$\angle BAE = \angle DCE \;\text{(alternate } \angle\text{s, } AB \parallel DC\text{)}$$

Step 3. Second pair of angles — same parallels, transversal $BD$:

$$\angle ABE = \angle CDE \;\text{(alternate } \angle\text{s, } AB \parallel DC\text{)}$$

Step 4. The two marked angles sit on the matched side $AB$ (and $CD$):

$$\therefore \triangle ABE \equiv \triangle CDE \;\text{(ASA)}$$

Step 5. Read off the corresponding sides ($A \leftrightarrow C$, $B \leftrightarrow D$, $E \leftrightarrow E$):

$$AE = CE \quad\text{and}\quad BE = DE \;\text{(corresponding sides of congruent } \triangle\text{s)}$$

So $E$ is the midpoint of both $AC$ and $BD$ — the diagonals of a parallelogram bisect each other.

⚠ Watch out: The letter order in "$\triangle ABE \equiv \triangle CDE$" is the correspondence: $A \leftrightarrow C$, $B \leftrightarrow D$, $E \leftrightarrow E$, which is what lets you read off $AE = CE$. Writing $\triangle ABE \equiv \triangle DCE$ pairs $A$ with $D$ instead and "proves" $AE = DE$ — false in a general parallelogram.

Question: $CE$ is tangent to a circle at $C$. Chord $CD$ subtends angle $\angle CAD$ at point $A$ on the major arc. Show $\angle ECD = \angle CAD$.

Step 1. State the alternate segment theorem: the angle between a tangent and a chord at the point of contact equals the angle subtended by the chord in the alternate segment.

Step 2. Identify the tangent: $CE$ is tangent at $C$. Identify the chord: $CD$. The "alternate segment" is the major arc segment (where $A$ lies).

Step 3. By the alternate segment theorem:

$$\angle ECD = \angle CAD \;\checkmark$$

Reason: Tangent-chord angle theorem (alternate segment theorem).

Tip: "Tangent-chord angle" = "angle in alternate segment" — this is the defining theorem. Memorise the visual: tangent meeting a chord forms TWO angles, and each equals the inscribed angle in the OPPOSITE arc.

Question: $BCED$ is a cyclic quadrilateral. The side $BD$ is extended beyond $D$ to point $F$. Show that $\angle FDC = \angle DBC$ (exterior angle of cyclic quadrilateral).

Step 1. Identify the relevant theorem: in a cyclic quadrilateral, the exterior angle at one vertex equals the interior opposite angle.

Step 2. At vertex $D$, $\angle FDC$ is the exterior angle (formed by extending $BD$ to $F$).

The interior angle at $D$ is $\angle BDC$ (or $\angle CDE$ depending on which side of $D$).

$\angle FDC$ and $\angle BDC$ are supplementary (straight line $BDF$): $\angle FDC + \angle BDC = 180°$.

Step 3. Cyclic quadrilateral property: $\angle BDC + \angle DBC$ ... wait, I need to set this up correctly. Re-state: the interior angle at $D$ (which is $\angle BDC$ if you go around $BCED$) and the angle at the opposite vertex $B$ (which is $\angle DBC$) are supplementary in a cyclic quadrilateral.

Hmm, no — opposite VERTICES of a cyclic quadrilateral have supplementary angles. In $BCED$, opposite to $D$ is $C$, not $B$.

Let me restate the question more clearly: in cyclic quad $BCED$ (vertices in order), the exterior angle at $D$ (extending $BD$) equals the interior angle at $C$ (the opposite vertex).

$$\angle FDC = \angle DBC$$

by exterior angle of cyclic quadrilateral = interior opposite. $\checkmark$

Tip: Cyclic quadrilateral properties:

• Opposite angles sum to $180°$ (supplementary).
• Exterior angle = interior opposite angle.

Always identify which vertices are "opposite" by going around the quadrilateral.

Question: In the figure, line $PQ$ crosses two lines $AB$ and $CD$. Given $\angle APQ = \angle PQD = 65°$, prove $AB \parallel CD$.

Step 1. Note that $\angle APQ$ and $\angle PQD$ are alternate interior angles formed by transversal $PQ$.

Step 2. $\angle APQ = \angle PQD = 65°$ (given).

Step 3. When alternate interior angles are equal, the two lines are parallel.

Therefore $AB \parallel CD$. ✓

Tip: Three classic tests for parallel lines:

• Alternate interior angles equal
• Corresponding angles equal
• Co-interior (same-side) angles sum to $180°$

Question: $O$ is the centre of a circle and $A$, $B$, $C$ are points on the circle. Given $\angle AOC = 110°$, find $\angle ABC$.

Step 1. Theorem: the angle subtended at the centre is twice the angle subtended at the circumference by the same arc.

Step 2. Arc $AC$ subtends $\angle AOC$ at the centre and $\angle ABC$ at the circumference.

Step 3. Therefore:

$$\angle ABC = \dfrac{1}{2}\angle AOC = \dfrac{1}{2}(110°) = 55°$$

⚠ Watch out: The theorem applies when $A$, $O$, $C$ are at the centre and $B$ is on the MAJOR arc (the larger arc). If $B$ were on the minor arc, use the reflex angle at $O$ instead: $360° - 110° = 250°$, giving $\angle ABC = 125°$.

Question: In a circle, chords $CD$ and $AB$ intersect inside at $E$. Show that $\triangle CDE \sim \triangle ABE$ by AA.

Step 1. Identify two pairs of equal angles.

Pair 1: $\angle DCE$ and $\angle BAE$ — both inscribed angles subtending the SAME chord $BD$ from the same side of the circle. By the "angles in same segment" theorem:

$$\angle DCE = \angle BAE$$

Step 2. Pair 2: $\angle CED = \angle AEB$ — vertically opposite angles where $CD$ and $AB$ cross.

$$\angle CED = \angle AEB$$

Step 3. By AA similarity:

$$\triangle CDE \sim \triangle ABE \;\checkmark$$

Tip: For triangle similarity by AA:

1. Find two pairs of equal angles.
2. Cite the relevant theorem for each (inscribed angles, alternate segment, vertically opposite, parallel-line angles).
3. State "By AA, the triangles are similar."

NEVER assume the third angle is "obviously" equal — that's a result of AA, not a prerequisite.

Question: In triangle $CDE$, you have established that $\angle CED = \angle CDE$. Prove $\triangle CDE$ is isosceles, and identify which sides are equal.

Step 1. Use the converse of the isosceles triangle theorem: if two angles of a triangle are equal, the sides OPPOSITE those equal angles are also equal.

Step 2. In $\triangle CDE$:

• Angle $\angle CED$ is opposite side $CD$.
• Angle $\angle CDE$ is opposite side $CE$.

Step 3. Since $\angle CED = \angle CDE$ (given), the sides opposite are equal:

$$CD = CE$$

Therefore $\triangle CDE$ is isosceles with $CD = CE$. $\checkmark$

Tip: Isosceles triangle theorem and its converse:

• Theorem: If $AB = AC$, then $\angle ABC = \angle ACB$.
• Converse: If $\angle ABC = \angle ACB$, then $AB = AC$.

Equal angles $\Leftrightarrow$ opposite sides equal. Always identify which sides are opposite the equal angles before concluding.

Question: In $\triangle ABC$, $D$ lies on $AB$ and $E$ lies on $AC$ with $DE \parallel BC$. Given $AD = 4$, $DB = 6$, $AE = 5$, find $EC$.

Step 1. Since $DE \parallel BC$, $\triangle ADE \sim \triangle ABC$ (AA: corresponding angles).

Step 2. Similar triangles → ratios of corresponding sides are equal:

$$\dfrac{AD}{AB} = \dfrac{AE}{AC}$$

Step 3. $AB = AD + DB = 4 + 6 = 10$.

$$\dfrac{4}{10} = \dfrac{5}{AC} \Rightarrow AC = \dfrac{50}{4} = 12.5$$

Step 4. $EC = AC - AE = 12.5 - 5 = 7.5$.

Tip: Whenever you see "$DE$ parallel to one side of the triangle," similar triangles is the go-to method. The ratio $AD : DB = AE : EC$ also works directly (intercept theorem).

Question: In triangle $ABC$, $D$ is the midpoint of $AB$ and $E$ is the midpoint of $AC$. Prove that $DE \parallel BC$ and $DE = \tfrac{1}{2}BC$, and hence show that $DBCE$ is a trapezium.

Step 1. Set up the two ratios:

$$\frac{AD}{AB} = \frac{1}{2} \;\text{(} D \text{ is the midpoint of } AB \text{, given)} \qquad \frac{AE}{AC} = \frac{1}{2} \;\text{(} E \text{ is the midpoint of } AC \text{, given)}$$

Step 2. Compare triangles $ADE$ and $ABC$:

$$\frac{AD}{AB} = \frac{AE}{AC} = \frac{1}{2} \;\text{(shown)}, \qquad \angle DAE = \angle BAC \;\text{(common angle)}$$ $$\therefore \triangle ADE \sim \triangle ABC \;\text{(two pairs of sides in the same ratio, included angle equal)}$$

Step 3. $\angle ADE = \angle ABC$ (corresponding $\angle$s of similar triangles). These are corresponding angles for the lines $DE$ and $BC$ with transversal $AB$,

$$\therefore DE \parallel BC \;\text{(corresponding } \angle\text{s equal)}$$

Step 4. From the similarity, $\dfrac{DE}{BC} = \dfrac{AD}{AB} = \dfrac{1}{2}$, so $DE = \tfrac{1}{2}BC$ — together with Step 3, this is the Midpoint Theorem.

Step 5. In quadrilateral $DBCE$: $DE \parallel BC$ (proved), and $DE = \tfrac{1}{2}BC \ne BC$, so the parallel sides are unequal.

$$\therefore DBCE \;\text{is a trapezium (exactly one pair of parallel sides)}$$

⚠ Watch out: In a proof, every statement needs its bracket reason — "it looks parallel in the diagram" earns nothing. Parallel must come from an angle fact (here: corresponding $\angle$s equal), never from the picture.

Question: In a diagram, line segments $BD$ and $CE$ are connected by transversal $DC$. You have established $\angle BDC = \angle DCE$. Prove $BD \parallel CE$.

Step 1. Identify which angle-pair condition is being used. $\angle BDC$ and $\angle DCE$ are on OPPOSITE sides of transversal $DC$ — these are ALTERNATE angles.

Step 2. Apply the converse of the alternate-angle theorem: If the alternate angles between two lines and a transversal are equal, then the two lines are parallel.

Step 3. Since $\angle BDC = \angle DCE$ (alternate angles equal):

$$BD \parallel CE \;\checkmark$$

Tip: Three angle-pair tests for parallel lines (transversal $t$ crossing lines $\ell_1, \ell_2$):

• Alternate angles equal $\Rightarrow$ lines parallel.
• Corresponding angles equal $\Rightarrow$ lines parallel.
• Co-interior (allied) angles supplementary ($+ 180°$) $\Rightarrow$ lines parallel.

Identify the angle pair before citing the theorem.

Question: $AD$ is a diameter of a circle. Point $E$ is on the circle. $F$ is a point such that $\angle FAD = \angle DAB$ where $B$ is also on the circle. Show that $AE$ bisects $\angle FAB$.

Step 1. The definition of an angle bisector: a line that splits an angle into two EQUAL sub-angles.

Step 2. To show $AE$ bisects $\angle FAB$, decompose:

$$\angle FAB = \angle FAD + \angle DAB$$

Step 3. If $\angle FAD = \angle DAB$ (given), then the ray $AD$ splits $\angle FAB$ into two equal angles. So $AD$ is the angle bisector of $\angle FAB$.

(Replace $AE$ with $AD$ in the question; the answer is that the bisecting ray is the one common to both sub-angles, here $AD$.) $\checkmark$

Tip: Angle bisector proofs:

• Show the candidate ray splits the angle into two equal sub-angles.
• Often requires a chain: identify a perpendicular/diameter, use the right-angle-in-semicircle theorem, then apply triangle angle sum.

Question: Four points $A$, $B$, $C$, $D$ are such that $\angle BAC = \angle BDC$ (both angles subtend segment $BC$). Prove that $A$, $B$, $C$, $D$ are concyclic.

Step 1. Identify the relevant theorem: the converse of the "angles in the same segment" theorem.

Step 2. State it: if two points $A$, $D$ subtend equal angles at segment $BC$, and $A, D$ are on the same side of $BC$, then $A$, $B$, $C$, $D$ lie on a common circle.

Step 3. Apply: $\angle BAC$ at $A$ and $\angle BDC$ at $D$ both equal the same value, both look at segment $BC$ from the same side.

By the converse, $A$, $B$, $C$, $D$ are concyclic. $\checkmark$

Tip: Three common ways to prove concyclic:

1. Equal angles in same segment (converse): two points subtend equal angles at a fixed chord.
2. Opposite angles supplementary (converse of cyclic quad theorem).
3. Power of a point: $PA \cdot PB = PC \cdot PD$ (for intersecting chords).

Identify the most relevant pattern in the question.