Chapter 11 — Triangles, a.k.a. Geometry
Nearly all of geometry comes down to the simple three-sided figure, the triangle. This is the longest and most important chapter — take your time and try problems after each section.
Topics in this chapter:
11.1 Classifying Triangles
The points where the sides of a triangle meet are the vertices. The sum of the three angles is always $180°$.
By Angles
| Type | Description |
|---|---|
| Acute | All three angles are acute ($< 90°$) |
| Right | One angle is exactly $90°$; the other two are complementary. The side opposite the right angle is the hypotenuse; the other sides are legs. |
| Obtuse | One angle is obtuse ($> 90°$) |
By Sides
| Type | Description |
|---|---|
| Equilateral | All three sides equal $\implies$ all angles equal $60°$ |
| Isosceles | Two sides equal (legs); the third is the base. The angles opposite the equal sides (base angles) are equal. The angle opposite the base is the vertex angle. |
| Scalene | No two sides are equal |
11.2 Parts of a Triangle
The sides of $\triangle ABC$ are usually called $a = BC$, $b = AC$, $c = AB$ (each side is named by the opposite vertex). The perimeter is $p = a + b + c$ and the semiperimeter is $s = p/2$.
Medians and the Centroid
A median is a segment from a vertex to the midpoint of the opposite side. The three medians are concurrent at the centroid $G$.
The centroid divides each median in a $2:1$ ratio from vertex to midpoint:
\[\frac{AG}{GD} = \frac{BG}{GE} = \frac{CG}{GF} = 2\]Angle Bisectors and the Incircle
An angle bisector divides an angle into two equal angles. It consists of all points equidistant from the two sides of the angle.
The three angle bisectors are concurrent at the incenter $I$. The incenter is equidistant from all three sides; this common distance is the inradius $r$.
The incircle is the circle with center $I$ and radius $r$, tangent to all three sides.
Perpendicular Bisectors and the Circumcircle
A perpendicular bisector of a segment is the line perpendicular to the segment through its midpoint. It consists of all points equidistant from the segment’s endpoints.
The three perpendicular bisectors are concurrent at the circumcenter $O$. The circumcircle passes through all three vertices with circumradius $R$.
- Acute triangle: circumcenter inside the triangle.
- Right triangle: circumcenter at the midpoint of the hypotenuse.
- Obtuse triangle: circumcenter outside the triangle.
Altitudes and the Orthocenter
An altitude is a perpendicular segment from a vertex to the line containing the opposite side. The altitudes are denoted $h_a$, $h_b$, $h_c$.
The three altitudes are concurrent at the orthocenter $H$.
- Acute triangle: orthocenter inside.
- Right triangle: orthocenter at the vertex of the right angle.
- Obtuse triangle: orthocenter outside.
Summary of Triangle Centers:
| Center | Construction | Property |
|---|---|---|
| Centroid $G$ | Medians | Divides medians $2:1$ |
| Incenter $I$ | Angle bisectors | Equidistant from all sides (inradius $r$) |
| Circumcenter $O$ | Perpendicular bisectors | Equidistant from all vertices (circumradius $R$) |
| Orthocenter $H$ | Altitudes | — |
Example 11-1. Show that the circumcenter of a right triangle is the midpoint of its hypotenuse.
Proof: Since $\angle C$ is an inscribed right angle, $\overset{\frown}{AB} = 2 \angle C = 180°$. So $AB$ is a diameter, and its midpoint $O$ is the center of the circumscribed circle.
Exercise 11-1. Show that the circumradius of a right triangle equals half the hypotenuse.
Exercise 11-2. Show that the median to the hypotenuse of a right triangle equals half the hypotenuse.
Exercise 11-3. Show that if a median of a triangle equals half the side to which it is drawn, the triangle is right.
11.3 The Triangle Inequality
Triangle Inequality. Given two sides of a triangle, the third side must be less than the sum and greater than the absolute difference of the first two:
\[|a - b| < c < a + b\]If $c = a + b$, the triangle is degenerate (a straight line).
Example 11-2. If two sides of a nondegenerate triangle are $7$ and $13$, what are the restrictions on the third side $x$?
Solution: We need:
- $x + 7 > 13 \implies x > 6$
- $7 + 13 > x \implies x < 20$
Thus $6 < x < 20$.
Exercise 11-4. In how many ways can we form a nondegenerate triangle by choosing three distinct numbers from ${1, 2, 3, 4, 5}$ as the sides?
11.4 The Pythagorean Theorem
Pythagorean Theorem. In a right triangle with legs $a$, $b$ and hypotenuse $c$:
\[a^2 + b^2 = c^2\]Example 11-3. Legs of a right triangle are $8$ and $4$. The hypotenuse is $\sqrt{8^2 + 4^2} = \sqrt{80} = 4\sqrt{5}$.
Example 11-4. In $\triangle ABC$, $\angle A + \angle B = 90°$, $AC = 4$, $AB = 5$. Find $BC$.
Solution: $\angle C = 90°$. By the Pythagorean Theorem: $4^2 + BC^2 = 5^2 \implies BC = 3$.
Example 11-5. Show that $BX + XC = BC$ if and only if $X$ is on segment $BC$.
Proof sketch: If $X$ is not on $BC$, drop a perpendicular from $X$ to $BC$ at $Y$. By the Pythagorean Theorem, $XB = \sqrt{XY^2 + BY^2} > BY$, and similarly $XC > YC$. So $XB + XC > BY + YC = BC$.
Example 11-6. Show that if $a$, $b$, $c$ are sides of an obtuse triangle with $a < b < c$, then $a^2 + b^2 < c^2$.
Proof: Draw altitude $h$ from $A$ to $\overline{BC}$, creating foot $D$ with $BD = a + x$. From right triangles: $c^2 = (a+x)^2 + h^2 = a^2 + (h^2 + x^2) + 2ax = a^2 + b^2 + 2ax > a^2 + b^2$.
Pythagorean Triples
A Pythagorean triple is a set of positive integers $(a, b, c)$ with $a^2 + b^2 = c^2$.
If $(a, b, c)$ is a triple, then $(na, nb, nc)$ is also a triple for any positive $n$.
Common Pythagorean triples:
| Triple | Multiples |
|---|---|
| $(3, 4, 5)$ | $(6, 8, 10)$, $(9, 12, 15)$, $(12, 16, 20)$, … |
| $(5, 12, 13)$ | $(10, 24, 26)$, … |
| $(7, 24, 25)$ | $(14, 48, 50)$, … |
| $(8, 15, 17)$ | $(16, 30, 34)$, … |
Example 11-7. The legs of a right triangle are $\frac{3}{105}$ and $\frac{4}{105}$. Find the hypotenuse.
Solution: Ratio is $3:4$, so the triple is $(3,4,5)$. Hypotenuse $= \frac{5}{105} = \frac{1}{21}$.
Example 11-8. Hypotenuse $= 4.25$, one leg $= 2$. Find the other leg.
Solution: Ratio $2 : 4.25 = 8 : 17$. The triple $(8, 15, 17)$ gives the other leg $= \frac{15}{4} = 3.75$.
Exercise 11-5. Find the altitude to the base of an isosceles triangle with base $16$ and legs $10$.
Exercise 11-6. How many non-congruent obtuse triangles with integer sides have perimeter $11$?
Exercise 11-7. Show that if $a$, $b$, $c$ are sides of an acute triangle, then $a^2 + b^2 > c^2$.
Exercise 11-8. A 25-foot ladder is placed against a wall, 7 feet from the base. If the top slips 4 feet, how far does the foot slide? (MAΘ 1992)
Exercise 11-9. Find the hypotenuse of a right triangle with legs $9\sqrt{2}$ and $12\sqrt{2}$.
Exercise 11-10. Find the second leg of a right triangle with hypotenuse $175$ and one leg $49$.
11.5 Congruent Triangles
Two triangles are congruent ($\triangle ABC \cong \triangle DEF$) if all corresponding sides and angles are equal. We always list vertices in corresponding order.
Congruency Theorems
| Criterion | What to show |
|---|---|
| SSS | Three sides equal |
| SAS | Two sides and the included angle equal |
| ASA | Two angles and the included side equal |
| AAS | Two angles and a non-included side equal |
| HL | (Right triangles only) Hypotenuse and one leg equal |
| LL | (Right triangles only) Both legs equal (special case of SAS) |
| SA | (Right triangles only) One acute angle and any corresponding side equal |
Warning: There is no SSA congruency for general triangles. Two sides and a non-included angle do not guarantee congruence.
Example 11-9. $\triangle ABC \cong \triangle BAD$. Find $\angle D$.
Solution: $\angle ABC = \angle BAD = 60°$. Then $\angle D = \angle C = 180° - \angle BAC - \angle ABC = 50°$.
Example 11-10. Prove: if two angles of a triangle are equal, the sides opposite them are equal.
Proof: Draw altitude $AX$ from the vertex angle. In right triangles $AXB$ and $AXC$: $AX = AX$ and $\angle B = \angle C$. By SA, $\triangle AXB \cong \triangle AXC$, so $AB = AC$.
Exercise 11-11. Prove that a chord and a radius are perpendicular if and only if the radius bisects the chord.
Exercise 11-12. Given chords $AB$ and $CD$ with $AB = CD$, show that minor arcs $AB$ and $CD$ are equal.
Exercise 11-13. If arcs $AB$ and $CD$ of a circle are equal, show that segments $AB = CD$.
Exercise 11-14. If two sides of a triangle are equal, prove the angles opposite them are equal.
Exercise 11-15. Show that in an isosceles triangle the centroid, incenter, orthocenter, and circumcenter are collinear, and in an equilateral triangle they coincide.
11.6 Similar Triangles
Two triangles are similar ($\triangle ABC \sim \triangle DEF$) if one is a scaled version of the other. Corresponding sides have a constant ratio $k$, and corresponding angles are equal.
If the ratio of sides is $k$, the ratio of areas is $k^2$. All corresponding lengths (medians, altitudes, etc.) also have ratio $k$.
Similarity Theorems
| Criterion | What to show |
|---|---|
| AA | Two pairs of corresponding angles are equal (most common method) |
| SAS | Two sides have the same ratio and the included angles are equal |
| SSS | All three pairs of sides have the same ratio |
Warning: Equal corresponding angles implies similarity only for triangles, not for other polygons.
Example 11-11. On sides $AB$ and $AC$ of $\triangle ABC$, points $D$ and $E$ are chosen so that $DE \parallel BC$. If $AB = 3 \cdot AD$ and $DE = 6$, find $BC$.
Solution: $DE \parallel BC$ gives $\angle ADE = \angle ABC$ and $\angle AED = \angle ACB$. So $\triangle ADE \sim \triangle ABC$. Thus $BC/DE = AB/AD = 3$, so $BC = 18$.
Example 11-12. The altitude to the hypotenuse divides it into segments of $4$ and $8$. Find the altitude.
Solution: The altitude to the hypotenuse of a right triangle creates two triangles similar to each other and to the original. From $\triangle ADC \sim \triangle CDB$:
\[\frac{AD}{CD} = \frac{CD}{BD} \implies CD^2 = 4 \times 8 = 32 \implies CD = 4\sqrt{2}\]Key fact. Whenever you see an altitude to the hypotenuse of a right triangle, remember:
\[\triangle ACD \sim \triangle ABC \sim \triangle CBD\]Example 11-13. Angle Bisector Theorem. If $AX$ bisects $\angle A$ of $\triangle ABC$, then:
\[\frac{AC}{CX} = \frac{AB}{BX}\]Proof: Extend $AX$ to $E$ so that $BE \parallel AC$. Alternate interior angles give $\angle CAE = \angle AEB$. Since $AX$ bisects $\angle A$, $\angle EAB = \angle AEB$, so $AB = BE$. Then $\triangle BXE \sim \triangle CXA$ by AA, giving $AC/CX = BE/BX = AB/BX$.
Example 11-14. If $AX$ and $BY$ are angle bisectors meeting at incenter $I$, show that $\frac{AI}{IX} = \frac{AC}{CX}$.
Proof: $CI$ bisects $\angle C$. Apply the Angle Bisector Theorem to $\angle C$ of $\triangle ACX$: $AC/AI = CX/XI$, which rearranges to $AI/IX = AC/CX$.
Example 11-15. $AD = DB = 5$, $EC = 8$, $AE = 4$, $\angle AED = 90°$. Find $BC$. (MAΘ 1987)
Solution: Draw $BH \parallel DE$. Then $\triangle DAE \sim \triangle BAH$ with ratio $1:2$. So $AH = 8$, $EH = 4$, $HC = 4$. Pythagorean Theorem: $DE = 3$, $BH = 6$. Then $BC = \sqrt{36 + 16} = 2\sqrt{13}$.
Exercise 11-16. Chord $EF$ is the perpendicular bisector of chord $BC$, intersecting at $M$. Point $U$ is between $B$ and $M$, and $EU$ extended meets the circle at $A$. Which triangle is always similar to $\triangle EUM$? (AHSME 1963)
Exercise 11-17. In $\triangle ABC$, $M$ and $N$ are midpoints of $AB$ and $AC$. If $AB = 5$, $BC = 6$, $AC = 7$, find $MN$.
Exercise 11-18. In trapezoid $TAPZ$, $TZ \parallel AP \parallel ER$, and $R$, $E$ are midpoints of $AT$ and $PZ$. If $AP = 64$, $TZ = 28$, $AZ = 46$, find $OI$. (MAΘ 1990)
Exercise 11-19. Show that if $AB \parallel CD \parallel EF$, then $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$ where $x$, $y$, $z$ are the distances between the parallels in the appropriate diagram.
11.7 Introduction to Trigonometry
In right $\triangle ABC$ with $\angle C = 90°$:
| Function | Definition | Ratio |
|---|---|---|
| $\sin A$ | opposite / hypotenuse | $a/c$ |
| $\cos A$ | adjacent / hypotenuse | $b/c$ |
| $\tan A$ | opposite / adjacent | $a/b$ |
| $\csc A$ | $1/\sin A$ | $c/a$ |
| $\sec A$ | $1/\cos A$ | $c/b$ |
| $\cot A$ | $1/\tan A$ | $b/a$ |
Key Identities
Co-function identities:
\[\sin A = \cos(90° - A), \quad \tan A = \cot(90° - A), \quad \sec A = \csc(90° - A)\]Special Right Triangles
45-45-90 Triangle
An isosceles right triangle with legs $a$ has hypotenuse $a\sqrt{2}$.
\[\text{Side ratio: } 1 : 1 : \sqrt{2}\] \[\sin 45° = \cos 45° = \frac{\sqrt{2}}{2}, \quad \tan 45° = 1\]30-60-90 Triangle
In a $30°$-$60°$-$90°$ triangle, the hypotenuse is twice the shorter leg (opposite $30°$), and the longer leg is $\sqrt{3}$ times the shorter leg.
\[\text{Side ratio: } 1 : \sqrt{3} : 2\]Complete Trigonometric Values Table
| Function | $0°$ | $30°$ | $45°$ | $60°$ | $90°$ |
|---|---|---|---|---|---|
| $\sin$ | $0$ | $\frac{1}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{3}}{2}$ | $1$ |
| $\cos$ | $1$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{1}{2}$ | $0$ |
| $\tan$ | $0$ | $\frac{\sqrt{3}}{3}$ | $1$ | $\sqrt{3}$ | undef. |
Memory tip. If you forget whether $\sin 30° = 1/2$ or $\cos 30° = 1/2$, draw a 30-60-90 triangle. The shorter leg is opposite $30°$, so $\sin 30° = \frac{\text{short leg}}{\text{hypotenuse}} = \frac{1}{2}$.
Example 11-16. $\angle B = 90°$ and $\cot C = 5/6$ in $\triangle ABC$. Find $BC$ if $AC = 5\sqrt{61}$.
Solution: $\cot C = BC/AB = 5/6$. Let $BC = x$, so $AB = 6x/5$. Pythagorean Theorem:
\[x^2 + \frac{36x^2}{25} = \frac{61x^2}{25} = 25 \cdot 61 \implies x = 25\]Example 11-17. $AB = 8$, $AC = 8\sqrt{2}$, $\angle ABC = 45°$, $\angle ACB = 30°$. Find $BC$.
Solution: Draw altitude $AD$. $\triangle ABD$ is 45-45-90: $BD = 4\sqrt{2}$. $\triangle ACD$ is 30-60-90: $AD = 4\sqrt{2}$, $CD = 4\sqrt{6}$.
\[BC = BD + CD = 4\sqrt{2} + 4\sqrt{6}\]Exercise 11-20. In circle $O$ with radius $6$, $\overset{\frown}{AB} = 60°$ and $\overset{\frown}{CD} = 90°$. Find the difference in chord lengths $CD - AB$.
Exercise 11-21. Find the smallest positive angle $x$ (in degrees) such that $\sin 3x = \cos 7x$. (Mandelbrot #3)
Exercise 11-22. Find $AC$ in $\triangle ABC$ if $\angle A = 90°$, $\sec B = 4$, and $AB = 6$.
11.8 Area of a Triangle
Three fundamental area formulas:
| Formula | When to use |
|---|---|
| $[ABC] = \frac{1}{2} a \cdot h_a$ | Know a side and its altitude |
| $[ABC] = \frac{1}{2} ab \sin C$ | Know two sides and the included angle |
| $[ABC] = rs$ | Know the inradius and semiperimeter |
where $h_a$ is the altitude to side $a$, $r$ is the inradius, and $s$ is the semiperimeter.
Proof: $[ABC] = \frac{1}{2}ah_a$
For a right triangle, two congruent copies form a rectangle with area $a \times h_a$, so each triangle has area $\frac{1}{2}ah_a$. For a general triangle, draw the altitude to split it into two right triangles and sum.
Proof: $[ABC] = \frac{1}{2}ab\sin C$
Draw altitude $h_a$ from $A$. Then $\sin C = h_a / b$, so $h_a = b\sin C$. Substituting: $\frac{1}{2}a \cdot b\sin C$.
For obtuse angles, use $\sin(180° - \theta) = \sin\theta$.
Proof: $[ABC] = rs$
Connect the incenter $I$ to the vertices, splitting $\triangle ABC$ into three triangles $AIB$, $BIC$, $CIA$, each with altitude $r$:
\[[ABC] = \frac{rc}{2} + \frac{ra}{2} + \frac{rb}{2} = \frac{r(a+b+c)}{2} = rs\]Example 11-18. $AB = AC = 50$, $BC = 80$. Find $[ABC]$.
Solution: Altitude from $A$ bisects $BC$: $AX = \sqrt{50^2 - 40^2} = 30$. Area $= \frac{1}{2}(80)(30) = 1200$.
Example 11-19. Find the inradius of a triangle with perimeter $40$ and area $120$.
Solution: $[ABC] = rs = r \cdot 20 = 120 \implies r = 6$.
Example 11-20. Isosceles $\triangle ABC$ with $AB = AC = 4$, $\angle C = 75°$. Find the area.
Solution: $\angle B = 75°$, $\angle A = 30°$.
\[[ABC] = \frac{1}{2}(AB)(AC)\sin A = \frac{1}{2}(4)(4)\sin 30° = 4\]Example 11-22. Equilateral triangle with side $s$: find area.
Solution: $[ABC] = \frac{1}{2}s^2 \sin 60° = \frac{s^2\sqrt{3}}{4}$.
Example 11-23. Sides $8$, $15$, $17$. Find the inradius.
Solution: $8^2 + 15^2 = 17^2$ (right triangle). Area $= \frac{1}{2}(8)(15) = 60$. Semiperimeter $= 20$. Then $r = 60/20 = 3$.
Example 11-24. Proof of the Pythagorean Theorem (via areas).
Arrange four right triangles (legs $a$, $b$, hypotenuse $c$) inside a square of side $c$. The inner square has side $a - b$:
\[c^2 = (a-b)^2 + 4\cdot\frac{ab}{2} = a^2 - 2ab + b^2 + 2ab = a^2 + b^2\]Exercise 11-23. What is the area of an equilateral triangle with altitude $12$?
Exercise 11-25. Tangents from $C$ to circle $O$ are extended to $A$ and $B$ such that $AB$ is tangent to $O$ at $X$. If the perimeter of $\triangle ABC$ is $50$ and $[ABC] = 100$, find the area of circle $O$.
Exercise 11-26. Eight equally spaced points on a circle of radius $1$. Find the area of the regular octagon they form.
11.9 A Handful of Helpful Hints
Parallel and Perpendicular Lines
Parallel and perpendicular lines appear in three ways:
- Given — mark equal angles or right angles and proceed.
- Drew — adding an extra parallel or perpendicular line often simplifies the problem.
- Discovered — prove lines are parallel/perpendicular using angle relationships.
Proving Lines Parallel
Given two lines and a transversal, any of these implies the lines are parallel:
- A pair of alternate interior angles are equal.
- A pair of corresponding angles are equal.
- A pair of same-side interior angles are supplementary.
- Two points on one line are equidistant from the other line.
Proving Lines Perpendicular
Via angles:
- Show the angle between them is inscribed in a semicircle.
- Show adjacent angles formed are equal (and thus each $90°$).
Via right triangles:
- Two angles are complementary $\implies$ third is $90°$.
- Sides satisfy the Pythagorean Theorem.
- Similar/congruent to a known right triangle.
- A median equals half the side to which it is drawn.
Without angles:
- One line passes through the center of a circle tangent to the other at the intersection.
- A radius bisects a chord.
- One segment is an altitude to the other.
Example 11-25. In $\triangle ABC$, $AE:EB = 1:3$ and $CD:DB = 1:2$. Lines $AD$ and $CE$ intersect at $F$. Find $\frac{EF}{FC} + \frac{AF}{FD}$. (AHSME 1965)
Solution: Draw $DH \parallel EA$. Since $\triangle EBC \sim \triangle GDC$ with ratio $3:1$, $DG/EB = 1/3$. Since $EA/EB = 1/3$ also, $EA = DG$. By ASA, $\triangle EAF \cong \triangle GDF$, so $AF = FD$ and $AF/FD = 1$.
From $\triangle EBC \sim \triangle GDC$: $GC/EC = 1/3$. Then $EF = FG = GC = EC/3$, so $EF/FC = \frac{EC/3}{2EC/3} = \frac{1}{2}$.
\[\frac{EF}{FC} + \frac{AF}{FD} = \frac{1}{2} + 1 = \frac{3}{2}\]Exercises & Problems
167. In $\triangle ADC$, segment $DM$ is drawn such that $\angle ADM = \angle ACD$. Prove that $AD^2 = (AM)(AC)$.
168. How many scalene triangles with integral sides have perimeter less than $13$? (AHSME 1956)
169. Sides of $\triangle BAC$ are in ratio $2:3:4$. $BD$ is the angle bisector to the shortest side $AC$ (length $10$). Find the longer segment of $AC$. (AHSME 1966)
170. How many distinct lines represent the altitudes, medians, and angle bisectors of a non-equilateral isosceles triangle? (AHSME 1957)
171. $\triangle ABD$ is right at $B$. Point $C$ on $AD$ has $AC = CD$ and $AB = BC$. Find $\angle DAB$. (AHSME 1963)
172. Right triangle $PYT$ with $PY = 66$, $YT = 77$. If $PT > 50$ and $PT = x\sqrt{y}$ in simplified form, find $x + y$. (MAΘ 1990)
173. Triangle $PQR$ has sides $40$, $60$, $80$. The shortest altitude is $K$ times the longest altitude. Find $K$. (MATHCOUNTS 1990)
174. $\angle ACD = 90°$, $A$, $B$, $C$ collinear, $\angle A = 30°$, $\angle DBC = 45°$. If $AB = 3 - \sqrt{3}$, find $[\triangle BCD]$. (MAΘ 1992)
175. The perpendicular bisectors of two sides of $\triangle ABC$ meet the third side at the same point. Prove the triangle is right. (M&IQ 1992)
176. If $h_a$, $h_b$, $h_c$ are altitudes, show $\frac{1}{h_a} < \frac{1}{h_b} + \frac{1}{h_c}$.
177. Right $\triangle ABC$, $\angle C = 90°$, $\sin A = 7/25$. Find $\sin B$, $\cos A$, $\cot A$, $\csc B$.
178. Right $\triangle ABC$: the angle between median $CM$ and hypotenuse $AB$ is $30°$. Altitude $CH = 4$. Find the area. (M&IQ 1992)
179. Base of a triangle is 15 inches. Two lines parallel to the base divide the triangle into three equal areas. Find the length of the parallel closer to the base. (AHSME 1953)
180. $AB$ is divided at $C$ so that $AC = 3CB$. Circles on $AC$ and $CB$ as diameters; a common tangent meets $AB$ extended at $D$. Show $BD$ equals the radius of the smaller circle. (AHSME 1954)
181. Medians $AD$ and $BE$ of right $\triangle ABC$ (hypotenuse $AB$). A right triangle with legs $AD$ and $BE$: find its hypotenuse in terms of $AB$. (Mandelbrot #2)
182. $M$ is the midpoint of $AB$ in equilateral $\triangle ABC$. $N$ on $BC$ with $MN \perp BC$. Prove $4BN = BC$. (M&IQ 1992)
183. Right $\triangle ACD$ with $\angle D = 90°$, $B$ on $AD$, $AB = 1$. $\angle DAC = \alpha$, $\angle DBC = \beta$. Find $DC$ in terms of $\alpha$ and $\beta$. (MAΘ 1991)
184. $\angle B$ of $\triangle ABC$ is trisected by $BD$ and $BE$ meeting $AC$ at $D$ and $E$. Prove $\frac{AD}{EC} = \frac{(AB)(BD)}{(BE)(BC)}$. (AHSME 1952)
185. $I$ is the incenter of $\triangle ABC$ with $AB = AC = 5$, $BC = 8$. Find $AI$. (Mandelbrot #3)
186. Equilateral $\triangle ABC$ with $F$, $Q$, $N$ such that $AF = QB = NC = \frac{2}{3}AB$. Prove $\angle AFQ = \angle NQB = \angle FNC = 90°$ and $\triangle FQN$ is equilateral. (M&IQ 1992)
187. The area of a triangle equals the product of an altitude and the median to the same side. Prove the triangle is right. (M&IQ 1992)
188. $\triangle ABC$: $\angle A = 100°$, $\angle B = 50°$, $\angle C = 30°$. $AH$ altitude, $BM$ median. Find $\angle MHC$. (AHSME 1989)
189. Two altitudes of a scalene triangle have length $4$ and $12$. If the third is an integer, what is the maximum? (AHSME 1986)
190. Medians from the acute vertices of a right triangle are $5$ and $\sqrt{40}$. Find the hypotenuse. (AHSME 1951)
191. If $\tan x = \frac{2ab}{a^2 - b^2}$ where $a > b > 0$ and $0 < x < 90°$, find $\sin x$ in terms of $a$ and $b$. (AHSME 1972)
192. Right $\triangle ABC$ with $F$ midpoint of hypotenuse $AB$ and $BC = 3AC$. $D$ and $E$ divide $BC$ into thirds. Prove $\triangle DFE$ is isosceles right. (M&IQ 1992)
193. Median to a 10 cm side has length 9 cm and is perpendicular to the second median. Find the third median. (MAΘ 1990)
194. A point inside an equilateral triangle: the sum of the perpendicular distances to the three sides equals the altitude length. (AHSME 1950)
195. $M$ midpoint of $AB$ in equilateral $\triangle ABC$. $N$, $S$, $K$ divide $BC$ into four equal segments. $P$ midpoint of $CM$. Prove $\angle MNB = \angle KPN = 90°$. (M&IQ 1992)
196. Prove: in $\triangle ABC$, if $\angle A > \angle B$, then $BC > AC$.