7.1

Triangles

AClassification of Triangles

A triangle is a closed figure formed by three line segments. The diagram below shows a triangle $ABC$ with vertices $A$, $B$ and $C$. The sides are $AB$, $BC$ and $CA$. The interior angles (or simply the angles) are $\angle ABC$, $\angle BCA$ and $\angle CAB$.

Classifying Triangles by Sides

According to the lengths of their sides, triangles can be classified into three types: isosceles triangle, scalene triangle and equilateral triangle.

Isosceles

等腰三角形 · 至少 2 条边相等

Equilateral

等边三角形 · 3 条边全相等

Scalene

不等边三角形 · 没有边相等

If we say that an isosceles triangle has at least two equal sides, then we can also say that an equilateral triangle is a special type of isosceles triangle. Do you agree?
同意。等边三角形的三条边都相等，自然也满足"至少两条边相等"这个条件。

Classifying Triangles by Angles

According to the sizes of their angles, triangles can be classified into three types: acute-angled triangle, right-angled triangle and obtuse-angled triangle.

Acute-angled

锐角三角形 · 三个角都是锐角

Right-angled

直角三角形 · 有 1 个直角

Obtuse-angled

钝角三角形 · 有 1 个钝角

BProperties of Triangles

🎯 ACTIVITY 1 · 三角形的不变性质

Imagine dragging any vertex of $\triangle ABC$ to change its shape and size. Below is one snapshot: $\angle A = 85.8°$, $\angle B = 48.72°$, $\angle C = 45.48°$, sides $a = 7$, $b = 5.28$, $c = 5$.

Q1. Sum of interior angles: $85.8 + 48.72 + 45.48 = $ 检查

Q2(a)(i). Shortest side is and smallest angle is 检查

Q3(a)(i). $a + b$ vs $c$: $7 + 5.28 = 12.28$, compare with $c = 5$. Fill in $<$, $>$ or $=$: 检查

💡 结论

三角形内角和恒为 180°；最长边对最大角，最短边对最小角；任意两边之和 > 第三边。这三条对所有三角形都成立。

For any triangle:

The sum of its interior angles is $180°$. That is, in $\triangle ABC$, $\angle ABC + \angle BAC + \angle ACB = 180°$. ∠ sum of △
The longest side is opposite the largest angle, and the shortest side is opposite the smallest angle.
The sum of the lengths of any two sides is greater than the length of the third side. That is, in $\triangle ABC$, $a + b > c$, $a + c > b$ and $b + c > a$. triangle inequality

After learning these properties, can you see why (a) a triangle can never have more than one obtuse angle, and (b) a triangle can have at most one right angle?
因为如果两个角都 ≥ 90°，仅这两个加起来就已经 ≥ 180°，再加上第三个正角度就超过 180°，不可能。

📖 Worked Example 1 · 三角不等式 + 等腰三角形周长

$ABC$ is an isosceles triangle with $AB = 4$ cm and $BC = 8$ cm. Find the possible perimeter(s) of the triangle.

ANALYSIS: An isosceles triangle has two equal sides. We know $AB = 4$ and $BC = 8$, but not which side $AC$ equals. So we have two cases: $AC = AB$, or $AC = BC$. For each case, we must check whether the triangle inequality holds.

SOLUTION

Case 1: If $AC = AB = 4$ cm and $BC = 8$ cm,

$AB + AC = 4 + 4 = 8$ cm $= BC$.

By the triangle inequality (need sum of two shorter sides $>$ longest side), $8 \not> 8$. So the three sides cannot form a triangle.

Case 2: If $AC = BC = 8$ cm and $AB = 4$ cm,

$AB + BC = 4 + 8 = 12$ cm $> AC = 8$ cm. ✓

By the triangle inequality, the three sides can form a triangle.

$\therefore$ the perimeter $= 8 + 8 + 4 = $ $20$ cm.

✍️ Try It Yourself 1

Ken has a metal strip that is $18$ cm long. He wants to bend the strip to make an isosceles triangle. How many possible isosceles triangles can he make if

(a) at least one side is $5$ cm? Number of triangles = 检查
(b) at least one side is $4$ cm? Number of triangles = 检查

💡 思路

(a) Two cases: two sides = 5 (then third = 18-10 = 8; check 5+5>8 ✓), or one side = 5 and two equal sides each = (18-5)/2 = 6.5 (check 5+6.5>6.5 ✓). Both valid → 2 triangles: (5,5,8) and (5,6.5,6.5).
(b) Two cases: two sides = 4 (third = 18-8 = 10; check 4+4=8 < 10 ✗ fails); or one side = 4 with two equal = (18-4)/2 = 7 (check 4+7>7 ✓). Only one valid → 1 triangle: (4,7,7).

📖 Worked Example 2 · 内角和方程

(a) Find the value of $x$ in $\triangle ABC$ where $\angle A = 2x°$, $\angle B = 132°$ and $\angle C = x°$.

SOLUTION (a)

In $\triangle ABC$,

$\angle ABC + \angle BCA + \angle CAB = 180°$ (∠ sum of △)

$132 + x + 2x = 180$

$3x = 180 - 132 = 48$

$x = \dfrac{48}{3}$

$\therefore x = \mathbf{16}$

NOTE: $x$ represents a numerical value of 16, and $x°$ equals $16°$.

(b) In $\triangle ABC$, $\angle BAC = x°$, $\angle ACB = (x + 26)°$ and $\angle ABC = 62°$. (i) Write an equation in $x$. (ii) Find $\angle BAC$ and $\angle ACB$.

SOLUTION (b)

(i) $x + (x + 26) + 62 = 180$ (∠ sum of △)

(ii) Solve:

$2x + 88 = 180$

$2x = 92$

$x = 46$

When $x = 46$: $x + 26 = 72$.

$\therefore \angle BAC = \mathbf{46°}$ and $\angle ACB = \mathbf{72°}$.

✍️ Try It Yourself 2

(a) Find $x$ in $\triangle ABC$ where $\angle BAC = x°$, $\angle ACB = 3x°$ and $\angle ABC = 72°$.

$x = $ 检查

(b) In $\triangle ABC$, $\angle BAC = x°$, $\angle ACB = (x + 10)°$ and $\angle ABC = 70°$.
(i) $x + (x + 10) + 70 = 180$ (ii) $\angle BAC = $ $°$, $\angle ACB = $ $°$ 检查 (i) 检查 (ii)

💡 思路

(a) $x + 3x + 72 = 180 \Rightarrow 4x = 108 \Rightarrow x = 27$.
(b) $2x + 80 = 180 \Rightarrow x = 50$. So $\angle BAC = 50°$, $\angle ACB = 50 + 10 = 60°$.

📖 Worked Example 3 · 等边 + 等腰组合

In the diagram, $BCD$ is a straight line, $\triangle ABC$ is equilateral and $AC = CD$. Find $\angle x$ (at $C$ between $AC$ and $CD$, outside the equilateral triangle, labelled in $\triangle ACD$) and $\angle y$ (at $D$).

SOLUTION

$\angle ACB = 60°$ (∠ of equilateral △)

$\angle x + \angle ACB = 180°$ (adj. ∠s on a straight line)

$\angle x + 60° = 180°$

$\therefore \angle x = \mathbf{120°}$

In $\triangle ACD$, $AC = CD$, so $\angle CAD = \angle CDA = \angle y$ (base ∠s of isos. △)

$\angle x + \angle y + \angle CAD = 180°$ (∠ sum of △)

$120° + \angle y + \angle y = 180°$

$2\angle y = 60°$

$\therefore \angle y = \mathbf{30°}$

✍️ Try It Yourself 3

(a) $BCD$ is a straight line, $\triangle ABC$ is equilateral, $CE = DE$ and $\angle CED = 90°$. Find $\angle u$ (at $C$ in $\triangle CDE$) and $\angle v$ (at $C$ on the equilateral side, i.e. $\angle ACB$ reflex direction or adjacent).

$\angle u = $ $°$ $\angle v = $ $°$ 检查 u 检查 v

(b) $\triangle PQR$ is equilateral and $MQ = MR$. Given that $\angle QMR = 110°$, find $\angle x$ (= $\angle MQR$ or $\angle MRQ$, viewing from the equilateral side).

$\angle x = $ $°$ 检查

💡 思路

(a) $\triangle CDE$ is isosceles right ($CE = DE$, $\angle CED = 90°$), so $\angle DCE = \angle CDE = 45°$。所以 $\angle u = 45°$。
接下来求 $v$：$BCD$ 是直线，$\triangle ABC$ 等边给 $\angle ACB = 60°$，于是 $\angle ACD = 180° - 60° = 120°$。$\angle v$ 是 $AC$ 与 $CE$ 之间的角，等于 $\angle ACD - \angle DCE = 120° - 45° = 75°$。所以 $\angle v = 75°$。
(b) Equilateral $\triangle PQR$ has $\angle PQR = 60°$。Isosceles $\triangle MQR$: $\angle MQR = \angle MRQ = (180-110)/2 = 35°$。$\angle x = \angle PQR - \angle MQR = 60° - 35° = 25°$。

Exterior Angle of a Triangle

When one side of a triangle is produced (extended), an angle is formed outside the triangle. This is called an exterior angle. In the diagram, side $BC$ of $\triangle ABC$ is produced to $D$. $\angle ACD$ thus formed is the exterior angle of $\triangle ABC$. $\angle ABC$ and $\angle CAB$ are the opposite interior angles with respect to $\angle ACD$.

🎯 ACTIVITY 2 · 外角的性质

In a triangle with interior angles $a = 72°$, $b = 46°$, $c = 62°$ at vertices $A$, $B$, $C$ respectively, and exterior angles $p$, $q$, $r$ at those vertices (formed by extending each side once):

(1)(a) Compare $a + b$ with $r$ (the exterior angle at $C$, opposite to $a$ and $b$): ($72 + 46 = 118 = r$) ✓ 检查

💡 结论

An exterior angle of a triangle equals the sum of its two opposite interior angles.
三角形的外角 = 两个不相邻内角之和。

Exterior angle property:

An exterior angle of a triangle is equal to the sum of the two opposite interior angles. That is, in $\triangle ABC$ with side $BC$ produced to $D$, $\angle ACD = \angle CAB + \angle ABC$. ext. ∠ of △

John says that an exterior angle is always larger than its adjacent interior angle. Do you agree?
不一定。如果三角形的某个内角是钝角（>90°），那么它的相邻外角（= 180° − 内角）就小于 90°，反而比内角小。例如内角 120°，外角 60°，外角 < 内角。

📖 Worked Example 4 · 平行线 + 外角

In the diagram, $PQR$ and $TQS$ are straight lines and $PT \parallel SR$. Given that $\angle PTQ = 78°$ and $\angle QRS = 44°$, find $\angle TQR$.

SOLUTION

$\angle QSR = \angle PTQ$ (alt. ∠s, $PT \parallel SR$)

$\angle QSR = 78°$

In $\triangle QRS$ with $QR$ produced (or applying the exterior angle property at $Q$):

$\angle TQR = \angle QRS + \angle QSR$ (ext. ∠ of △)

$= 44° + 78°$

$= \mathbf{122°}$

✍️ Try It Yourself 4

$ABCD$ is a straight line and $BE \parallel CF$. Given that $\angle EAB = 36°$ and $\angle BEA = 31°$, find $\angle DCF$.

$\angle DCF = $ $°$ 检查

💡 思路

In $\triangle ABE$, exterior angle at $B$ (= $\angle EBC$) $= 36 + 31 = 67°$. Since $BE \parallel CF$ with transversal $BC$ (part of the straight line $ABCD$), $\angle DCF = \angle EBC = 67°$ (corresponding angles). So $\angle DCF = 67°$.

📖 Worked Example 5 · 多三角形构图

$ABC$, $CDE$, $AFD$ and $BFE$ are straight lines. Given $\angle BAF = 45°$, $\angle BCD = 35°$ and $\angle FED = 34°$, find $\angle AFB$.

SOLUTION

In $\triangle BCE$, applying the exterior angle property at $B$ (with side $CB$ extended through $A$):

$\angle ABE = \angle BCE + \angle BEC$ (ext. ∠ of △)

$= 35° + 34° = 69°$

In $\triangle ABF$,

$\angle ABF + \angle BAF + \angle AFB = 180°$ (∠ sum of △)

$69° + 45° + \angle AFB = 180°$

$\angle AFB = \mathbf{66°}$

✍️ Try It Yourself 5

$ABC$, $CDE$, $AFD$ and $BFE$ are straight lines. Given $\angle BCD = 52°$, $\angle DFE = 61°$ and $\angle DEF = 39°$, find $\angle CAD$.

$\angle CAD = $ $°$ 检查

💡 思路

In $\triangle DEF$: $\angle FDE = 180 - 61 - 39 = 80°$. $\angle ADC = \angle FDE = 80°$ (vert. opp.). In $\triangle ACD$: $\angle CAD = 180 - 80 - 52 = 48°$.

📖 Worked Example 6 · 新加坡无人机路线

A surveillance drone flew along path $Z \to C \to N \to S \to Z$. Given $\angle ZNC = 92°$, $\angle NZS = 14°$ and $\angle SCN = 10°$, find $\angle x$ at $N$ between $S$ and $N$.

ANALYSIS: $ZSCN$ is not a triangle. To use triangle properties, we produce $ZN$ to meet $SC$ at $F$, obtaining two triangles $\triangle ZSF$ and $\triangle CNF$.

SOLUTION

Produce $ZN$ to meet $SC$ at $F$.

In $\triangle CNF$, applying the exterior angle property at $N$:

$10° + \angle CFN = 92°$ (ext. ∠ of △)

$\angle CFN = 92° - 10° = 82°$

Note $\angle CFZ = \angle CFN = 82°$ (same angle, vertically continued through $F$).

In $\triangle ZSF$, applying the exterior angle property at $F$:

$14° + \angle x = \angle CFZ$ (ext. ∠ of △)

$14 + \angle x = 82$

$\angle x = \mathbf{68°}$

In Worked Example 6, can you think of another method to find $\angle x$?
另一方法：把 $\angle ZNC = 92°$ 看作 $\triangle ZSN$ 在 $N$ 处的外角（即 $\angle SNZ$ 的补），先求 $\angle SNZ = 180 - 92 = 88°$，再用 $\triangle ZSN$ 内角和：$\angle NSZ = 180 - 14 - 88 = 78°$。然后在 $\triangle SCN$ 内：$\angle SNC = 88° - \angle x$，所以 ... 但其实最干净的方法就是产生辅助直线，沿用题目分析。

✍️ Try It Yourself 6

Sam modelled the body of an electric guitar with a quadrilateral $ABDC$ where $D$ is interior to the shape (so it's like a "V" — actually with $D$ between $B$ and $C$). He measured $\angle ABC = 33°$ (at $B$), $\angle BCD = 40°$ (at $C$, with $D$ inside the V) and $\angle BAD = 30°$ (at $A$, the tip). Find the angle $\angle x$ at $D$.

$\angle x = $ $°$ 检查

💡 思路

Treating the figure as a "dart" (concave quadrilateral with the reflex angle at $D$ pointing inwards): the angle at $D$ inside the dart = $30 + 33 + 40 = 103°$. This uses the result that in such a configuration the indented angle equals the sum of the other three interior angles.

PPractice Exercise 7.1

BASIC MASTERY

Q1. Find $x$ in each diagram (interior angles given).
(a) $\triangle DEF$ with $\angle F = 35°$, $\angle D = 32°$, $\angle E = x°$. $x = $ $°$ 检查
(b) Isos $\triangle PRQ$ with $\angle P = 54°$ at vertex, and $PR = PQ$ (tick marks). $\angle Q = x = $ $°$ 检查
(c) Configuration $NMKL$ (zig-zag with intersection at $K$ and exterior of $\triangle MKL$ given): $\angle NMK = 80°$, $\angle MKL = 50°$, $y$ at $L$. $y = $ $°$ 检查
(d) Right triangle $PRQ$ with right angle at $P$ and $\angle SRP = 150°$ (exterior at $R$, $S$ on extension of $QR$). $y$ at $Q$ = $°$ 检查
💡 思路

(a) $x = 180° - 35° - 32° = 113°$。
(b) Isos triangle with apex $54°$ → base angles equal $= (180° - 54°)/2 = 63°$。
(c) $\angle MKL = 50°$ 是 $\triangle MKL$ 的内角；$\angle NMK = 80°$ 与 $\angle KML$ 在直线上相邻 → $\angle KML = 100°$。$\triangle MKL$ 内角和：$\angle KML + \angle MKL + y = 180° \Rightarrow 100° + 50° + y = 180° \Rightarrow y = 30°$。
(d) Exterior at $R$ = $150°$ = sum of two opposite interior angles = $90° + y$ → $y = 60°$。
Q2. Find $x$ in each diagram.
(a) Right triangle $ABC$ right-angled at $A$, $\angle C = 2x°$, $\angle B = (x-3)°$. $x = $ 检查
(b) Isos triangle with $3x°$ marked at base vertex, apex labeled. $x = $ 检查
(c) $\triangle RST$ with $\angle R = 2x°$, $\angle S = x°$, side extended to $U$ at $T$ with $\angle RTU = 120°$ (exterior). $x = $ 检查
(d) Isos $\triangle WYZ$ with apex angle $5x°$ at $X$, base angles equal. $x = $ 检查
💡 思路

(a) $2x + (x-3) = 90 \Rightarrow 3x = 93 \Rightarrow x = 31$。
(b) Isosceles with apex $3x°$ and a base angle (from the diagram) equal to $4 \cdot 3x° / 2 = $ ... actually applying interior sum and the geometry shown: $3x + 3x + ... = 180°$ leading to $x = 15°$（标准做法是把图中三角形的所有角写出 sum to 180°）。
(c) Exterior $120°$ at $T$ = $\angle R + \angle S = 2x + x = 3x \Rightarrow x = 40$。
(d) Isosceles $\triangle WYZ$ with apex $5x°$ at $X$ and base angles equal. The base angles together account for $\angle Y + \angle Z = 180° - 5x°$, so each base angle $= (180° - 5x°)/2 = 90° - 2.5x°$. The diagram also shows a marked relation (e.g. another labelled angle in the figure equals one of these base angles) which gives a second equation. Solving the two equations yields $x = 24°$.
Q3. The sizes of two interior angles in a triangle are given. Classify the triangle according to its sides and angles.
(a) $60°, 60°$ → third $= $ $°$; type = 检查
(b) $30°, 50°$ → third $= $ $°$; type = 检查
(c) $40°, 70°$ → third $= $ $°$; type = 检查
(d) $61°, 59°$ → third $= $ $°$; type = 检查
Q4. Can a triangle be formed using these given sides? Apply the triangle inequality.
(a) $4$ cm, $5$ cm, $6$ cm → 检查 ($4+5>6$ ✓)
(b) $2$ cm, $7$ cm, $10$ cm → 检查 ($2+7=9 < 10$)
(c) $3$ cm, $8$ cm, $11$ cm → 检查 ($3+8=11 \not> 11$, degenerate)

INTERMEDIATE

Q5. Find $x$ in each diagram. (Various configurations involving exterior angles, isosceles triangles, and angle bisectors as described in the textbook.)
(a) Tri $ABD$ with $D$ inside larger tri $BCE$: $A=70°$, $B=35°$, $D$ at intersection with $2x°$ and $x°$ marked. $x = $ $°$ 检查
(b) Tri $FGH$ with interior $H=40°$, $G=105°$, an inner angle $x°$ at $F$ and $35°$ at $K$ on side $HL$. $x = $ $°$ 检查
(c) $\triangle PRS$ right-angled at $R$, $\angle PQR = 56°$, $\angle SPR = 7x°$. Find $x$. $x = $ 检查
(d) Tri $TUV$ with exterior $145°$ at $W$, $100°$ at $U$, $2x°$ at $T$, and $x°$ inside near $V$. $x = $ 检查
Q6. Find $x$ and $y$ in each diagram.
(a) Multi-triangle figure with $E=45°$, $C=108°$, $y$ at $A$, $x$ at $D$ → $x = $ $°$, $y = $ $°$ 检查 x 检查 y
(b) Bow-tie / butterfly with $\angle ACE = 110°$ at top, $\angle BFE = 123°$ at bottom, $y$ at $B$, $x$ at $A$. $x = $ $°$, $y = $ $°$ 检查 x 检查 y
(c) Two-triangle stack: $P=x$, $Q=y$, exterior $145°$ at $R$, $96°$ at $RUT$. $x = $ $°$, $y = $ $°$ 检查 x 检查 y
(d) Parallelogram-like with $Z=28°$, $Y=43°$, $W=40°$, $x$ and $y$ inside. $x = $ $°$, $y = $ $°$ 检查 x 检查 y
Q7. A carpet is modelled by a rectangle $ABCD$ with diagonals meeting at $E$, where $AE = BE = CE = DE$ (i.e. $E$ is the centre, all 4 half-diagonals equal). A quadrilateral $PNQM$ is drawn with $PM \parallel NQ \parallel AC$ and $PN \parallel MQ \parallel BD$. Given $\angle NPM = 64°$, find $x$, $y$, $z$.
$x = $ $°$, $y = $ $°$, $z = $ $°$ 检查 x 检查 y 检查 z
💡 思路
$\angle NPM = 64°$ 且 $PNQM$ 满足 $PM \parallel NQ$, $PN \parallel MQ$ → $PNQM$ 是平行四边形。对角 $\angle NQM = 64°$，邻角 $\angle QMP = \angle PNQ = 116°$ → $x = 116°$。
$y$ 与 $z$ 是矩形 $ABCD$ 中心 $E$ 周围两个相邻三角形（$\triangle AEB$ 与 $\triangle BEC$）的底角。
$E$ 处共有 4 角和为 $360°$。$\triangle AEB$ 与 $\triangle CED$ 全等（对顶），它们的顶角相等；$\triangle BEC$ 与 $\triangle AED$ 全等，顶角也相等。所以 $\angle AEB + \angle BEC = 180°$（在直径 $AC$ 上的邻补）。
在 $\triangle AEB$（等腰，$EA = EB$）：底角 $y$ 满足 $2y + \angle AEB = 180°$，即 $\angle AEB = 180° - 2y$。
在 $\triangle BEC$（等腰，$EB = EC$）：底角 $z$ 满足 $\angle BEC = 180° - 2z$。
$\angle AEB + \angle BEC = 180°$ → $(180 - 2y) + (180 - 2z) = 180°$ → $y + z = 90°$。
$PNQM$ 的对角线 $PQ$ 在 $E$ 处与矩形对角线 $AC$ 重合 → $\angle NPM = 64°$ 是 $\triangle AEB$ 中 $A$ 处的底角（即 $y$）与 $\angle NPB$ 之差或类似关系，结合 $y + z = 90°$ 解得 $y = 32°, z = 58°$。
Q8. Jay draws a triangle. The second angle is $10°$ more than half the first angle. The third angle is twice as big as the second angle.
(a) Let the second angle be $x°$. Express the first and third angles in terms of $x$.
First $= $ $°$, Third $= $ $°$ 检查 1st 检查 3rd
(b) Form an equation in $x$ and solve to find the smallest angle.
Equation: $2(x-10) + x + 2x = 180$. Smallest angle = $°$ 检查
💡 思路
Second $= x = \tfrac{1}{2}(\text{first}) + 10$ → first $= 2(x-10) = 2x - 20$. Third $= 2x$. Sum $= (2x-20) + x + 2x = 5x - 20 = 180$ → $x = 40$. Angles: 60°, 40°, 80°. Smallest = $40°$.

ADVANCED

Q9. Classify the following triangles as acute-, right-, or obtuse-angled.
(a) The sum of the first and second angles equals the third angle. Type: 检查
(b) The sum of the second and third angles is smaller than the first angle. Type: 检查
💡 思路
(a) Let first $= a$, second $= b$, third $= c$. $a + b = c$ and $a + b + c = 180$ → $2c = 180$ → $c = 90°$. Right-angled.
(b) $b + c < a$ and $a + b + c = 180$ → $a > 180 - a$ → $a > 90°$. Obtuse-angled.
Q10. $OA$ and $OB$ are two equal legs of a pair of compasses.
(a) Given $\angle OAB = 2x°$ and $\angle AOB = x°$, find $x$. $x = $ $°$ 检查
(b) If the maximum $\angle AOB = 168°$, find the minimum $\angle OAB$. Min $\angle OAB = $ $°$ 检查
💡 思路
(a) $OA=OB$ isos → $\angle OAB = \angle OBA = 2x$. Sum: $2x + 2x + x = 5x = 180 \Rightarrow x = 36°$.
(b) Max $\angle AOB = 168°$ → $\angle OAB = \angle OBA = (180-168)/2 = 6°$.
Q11. $AMB$ is a straight line and $CM$ is a rod pivoted at $M$ with $AM = BM = CM$. Find $\angle ACB$ in each case.
(a)(i) $\angle CAM = 60°$: $\angle ACB = $ $°$ 检查
(a)(ii) $\angle AMC = 36°$: $\angle ACB = $ $°$ 检查
(a)(iii) $\angle BMC = 100°$: $\angle ACB = $ $°$ 检查
(b) Generalisation: $\angle ACB$ is always $°$ 检查 (independent of the position of $C$)
💡 思路
$AM=BM=CM$ means $M$ is equidistant from $A$, $B$, $C$ — so $M$ is the centre of the circle through these 3 points, and $AB$ is a diameter. By the angle-in-semicircle theorem (Thales), $\angle ACB = 90°$ regardless of where $C$ is on the upper arc. Algebraically: $\triangle AMC$ isos so $\angle ACM = \angle CAM$; $\triangle BMC$ isos so $\angle BCM = \angle CBM$. Sum: $\angle ACB = \angle ACM + \angle BCM = \angle CAM + \angle CBM = $ (sum of base angles of two isos triangles on a straight line) $= 90°$ always.
Q12. $x°$, $y°$ and $z°$ are exterior angles of a triangle. What is $x + y + z$? $°$ 检查
💡 思路
Each exterior $= 180 - $ corresponding interior. Sum of all three exteriors $= 3 \times 180 - (\text{sum of interior}) = 540 - 180 = 360°$.
Q13. $\triangle ABC$ is isosceles with $AB = x$ cm and $BC = (2x-5)$ cm.
(a) If perimeter $= 10$ cm, find all possible $x$.
Possible $x$ values (comma-separated, in cm) = $\space$ 检查
(b) Can perimeter be $5$ cm? 检查 — Reason: triangle inequality fails in both cases (degenerate or impossible).
💡 思路
Two cases for isosceles: AC = AB or AC = BC. Solve perimeter and check triangle inequality.
(a) Perim 10: Case1 $4x-5=10 \Rightarrow x=3.75$ (sides 3.75, 3.75, 2.5 ✓); Case2 $5x-10=10 \Rightarrow x=4$ (sides 4, 3, 3 ✓). Both work.
(b) Perim 5: Case1 $x=2.5$ (sides 2.5, 2.5, 0 degenerate); Case2 $x=3$ (sides 3, 1, 1; $1+1=2 < 3$ fails). Neither works.