13.2 Right Triangles, Euclid's Altitude Laws, Angle Bisectors, Medians, Similarity & Area

13.2 Right Triangles, Euclid's Altitude Laws, Angle Bisectors, Medians, Similarity & Area

---

1. Right Triangles (Dik Üçgen), Triples & Euclid's Laws (Öklid)

Integer Pythagorean Triples & Rapid Radical Ratios

In any right triangle with legs $b, c$ and hypotenuse $a$, the Pythagorean theorem states $b^2 + c^2 = a^2$. Memorize the four standard primitive integer families and their scaled multiples ($k \in \mathbb{Z}^+$), plus two constant-ratio leg shortcuts:

• Primitive Triples: $(3k, 4k, 5k)$, $(5k, 12k, 13k)$, $(8k, 15k, 17k)$, and $(7k, 24k, 25k)$.
• $1 : 2 : \sqrt{5}$ Shortcut: If one leg is twice the other ($k$ and $2k$), the hypotenuse is immediately $k\sqrt{5}$.
• $1 : 3 : \sqrt{10}$ Shortcut: If one leg is three times the other ($k$ and $3k$), the hypotenuse is immediately $k\sqrt{10}$.

Special Angle Triangles (Özel Açılı Üçgenler)

• $30^\circ\text{-}60^\circ\text{-}90^\circ$ Triangle: Sides opposite $30^\circ, 60^\circ, 90^\circ$ are in the ratio $a : a\sqrt{3} : 2a$.
• $45^\circ\text{-}45^\circ\text{-}90^\circ$ Triangle: Isosceles right triangle with sides $a : a : a\sqrt{2}$.
• $15^\circ\text{-}75^\circ\text{-}90^\circ$ Triangle: If $h$ is the altitude dropped to the hypotenuse $c$, then the hypotenuse is four times the altitude:

$$h = \frac{c}{4} \iff c = 4h$$

Furthermore, if the leg opposite $15^\circ$ is $a$, the leg opposite $75^\circ$ is $a(2 + \sqrt{3})$.

• $30^\circ\text{-}30^\circ\text{-}120^\circ$ Obtuse Isosceles Triangle: If the two equal legs are $a$ and $a$, the side opposite $120^\circ$ is $a\sqrt{3}$.

Euclid's Right-Triangle Altitude Laws (Öklid Bağıntıları)

When $\triangle ABC$ has a right angle at $A$ ($m(\angle A) = 90^\circ$) AND an altitude $AH \perp BC$ of length $h$ that splits the hypotenuse $a = p + k$ into projection $p = \lvert BH \rvert$ (under leg $c$) and projection $k = \lvert HC \rvert$ (under leg $b$), four identities hold:

1. Altitude Law (Yükseklik Bağıntısı):

$$h^2 = p \cdot k$$

1. Leg Laws (Dik Kenar Bağıntıları):

$$c^2 = p \cdot (p + k) = p \cdot a \qquad \text{and} \qquad b^2 = k \cdot (p + k) = k \cdot a$$

1. Area / Product Law (Alan Bağıntısı):

$$b \cdot c = a \cdot h$$

---

2. Angle Bisectors (Açıortay) & Medians / Centroid (Kenarortay)

Interior & Exterior Angle Bisector Theorems

Every point on an angle bisector is equidistant from the two arms of the angle (dropping perpendiculars to both arms creates two congruent right triangles).

• Interior Angle Bisector Theorem (İç Açıortay Teoremi): In $\triangle ABC$, if the interior bisector $AN$ of $\angle A$ has length $n_A$ and divides base $BC$ into segments $m = \lvert BN \rvert$ (adjacent to leg $c$) and $n = \lvert NC \rvert$ (adjacent to leg $b$), then:

$$\frac{c}{b} = \frac{m}{n} \iff \frac{c}{m} = \frac{b}{n} \qquad \text{and} \qquad n_A^2 = b \cdot c - m \cdot n$$

• Exterior Angle Bisector Theorem (Dış Açıortay Teoremi): If the exterior bisector at vertex $A$ (with $c > b$) intersects the extension of base $BC$ (beyond $C$) at point $D$ at distance $d = \lvert CD \rvert$, then the "L-bounce" ratio from $D$ to the near corner $C$ vs. the far corner $B$ equals the near leg $b$ over the far leg $c$:

$$\frac{d}{d + a} = \frac{b}{c} \qquad \text{and} \qquad (n_A')^2 = d(d + a) - b \cdot c$$

Medians and the Centroid (Kenarortay ve Ağırlık Merkezi)

The three medians (kenarortaylar) of a triangle intersect at a single balance point $G$, the Centroid (Ağırlık Merkezi):

• $2 : 1$ Median Ratio: The centroid $G$ divides every median $AD$ in a $2 : 1$ ratio, with the longer segment adjacent to the vertex:

$$\lvert AG \rvert = 2\lvert GD \rvert \quad \text{(so total median } V_a = 3k = 2k + k\text{)}$$

• The $3\text{-}1\text{-}2$ Rule (3-1-2 Kuralı): If both the median $AD$ and the midline orta taban $EF$ (connecting the midpoints of $AB$ and $AC$) are drawn, $EF$ bisects $AD$ at $K$ ($3k$ above, $3k$ below) while $G$ divides $AD$ in $4k : 2k$. Thus, the segments along median $AD$ from vertex $A$ to $K$ to $G$ to $D$ are in the exact ratio:

$$\lvert AK \rvert : \lvert KG \rvert : \lvert GD \rvert = 3k : k : 2k$$

• Right-Triangle Centroid: In a right triangle ($m(\angle A) = 90^\circ$) with centroid $G$, combining Muhteşem Üçlü ($V_a = \frac{a}{2}$) with the $2:1$ centroid split gives $V_a = 3k \implies \text{hypotenuse } a = 6k$, where $\lvert AG \rvert = 2k$ and $\lvert GD \rvert = k$.

---

3. Triangle Similarity (Benzerlik) & Area (Üçgende Alan)

Parallel Similarity Patterns

1. Basic Proportionality (Temel Orantı Teoremi): If $DE \parallel BC$ cuts sides $AB$ and $AC$, then $\triangle ADE \sim \triangle ABC$ with similarity ratio:

$$\frac{\lvert AD \rvert}{\lvert AB \rvert} = \frac{\lvert AE \rvert}{\lvert AC \rvert} = \frac{\lvert DE \rvert}{\lvert BC \rvert} = k$$

When $D$ and $E$ are midpoints ($k = \frac{1}{2}$), $DE$ is the Midsegment (Orta Taban) and $\lvert DE \rvert = \frac{\lvert BC \rvert}{2}$. 2. Hourglass / Butterfly Similarity (Kelebek / Kum Saati Benzerliği): When $AB \parallel DC$ and diagonals $AC, BD$ intersect at $E$, $\triangle EAB \sim \triangle ECD$:

$$\frac{\lvert AB \rvert}{\lvert DC \rvert} = \frac{\lvert AE \rvert}{\lvert EC \rvert} = \frac{\lvert BE \rvert}{\lvert ED \rvert} = k$$

1. Angle-Angle Similarity (Açı-Açı Benzerliği): In overlapping or right-triangle puzzles without parallel lines, label two acute angles $\alpha$ and $\beta$ with $\alpha + \beta = 90^\circ$; any two triangles sharing acute angles $\alpha$ and $\beta$ are similar, and sides opposite equal angles are proportional.

Triangle Area Formulas & Partition Laws

• Base-Height, Sine, and Equilateral Area Formulas:

$$\text{Area}(\triangle ABC) = \frac{a \cdot h_a}{2} = \frac{1}{2} b \cdot c \cdot \sin A, \qquad \text{Equilateral Area} = \frac{a^2\sqrt{3}}{4} \quad \left(h = \frac{a\sqrt{3}}{2}\right)$$

• Equal-Height Base Ratio Law: If two triangles share the same vertex and their bases lie on the same line, they share the same altitude $h$, so their areas are directly proportional to their base lengths:

$$\frac{\text{Area}(\triangle ABD)}{\text{Area}(\triangle ADC)} = \frac{\lvert BD \rvert}{\lvert DC \rvert}$$

• Centroid Area Split: Drawing all three medians partitions $\triangle ABC$ into 6 equal-area triangles ($S, S, S, S, S, S$). Connecting $G$ only to the three vertices $A, B, C$ splits the area into 3 equal triangles ($2S, 2S, 2S$).
• Square-of-Similarity Area Law: If $\triangle_1 \sim \triangle_2$ with linear similarity ratio $k$, the ratio of their areas is $k^2$:

$$\frac{\text{Area}(\triangle_1)}{\text{Area}(\triangle_2)} = k^2$$

Consequently, equally spaced parallel lines dividing the legs into equal segments ($1 : 2 : 3 : 4$) produce cumulative areas $S, 4S, 9S, 16S$, so consecutive strips have areas $S, 3S, 5S, 7S, \dots$

---

Worked Examples

Worked Example 1: Euclid's Laws Combined with an Angle Bisector

Problem: In right triangle $\triangle ABC$ with $m(\angle BAC) = 90^\circ$, altitude $AH \perp BC$ divides hypotenuse $BC$ into $\lvert BH \rvert = 4\text{ cm}$ and $\lvert HC \rvert = 9\text{ cm}$. Calculate: (a) the altitude $h = \lvert AH \rvert$, (b) the legs $c = \lvert AB \rvert$ and $b = \lvert AC \rvert$, and (c) the segments into which the interior angle bisector of $\angle A$ divides hypotenuse $BC$.

Solution:

1. Altitude by Euclid's First Law: $h^2 = p \cdot k = 4 \cdot 9 = 36 \implies h = 6\text{ cm}$.
2. Legs by Euclid's Leg Laws: The full hypotenuse is $a = 4 + 9 = 13\text{ cm}$. Thus:

$$c = \sqrt{p \cdot a} = \sqrt{4 \cdot 13} = 2\sqrt{13}\text{ cm}, \qquad b = \sqrt{k \cdot a} = \sqrt{9 \cdot 13} = 3\sqrt{13}\text{ cm}$$

1. Angle Bisector Ratio: The interior angle bisector $AN$ of $\angle A$ divides $BC = 13\text{ cm}$ in the ratio of the adjacent legs:

$$\frac{\lvert BN \rvert}{\lvert NC \rvert} = \frac{c}{b} = \frac{2\sqrt{13}}{3\sqrt{13}} = \frac{2}{3}$$

Setting $\lvert BN \rvert = 2t$ and $\lvert NC \rvert = 3t$ with $2t + 3t = 13$ gives $t = \frac{13}{5}$, so $\lvert BN \rvert = \frac{26}{5} = 5.2\text{ cm}$ and $\lvert NC \rvert = \frac{39}{5} = 7.8\text{ cm}$.

Worked Example 2: Centroid 3-1-2 Rule and Area Partitioning

Problem: In $\triangle ABC$, $G$ is the centroid (ağırlık merkezi), $AD$ is the median to side $BC$, and $EF$ is the midsegment connecting the midpoints $E$ of $AB$ and $F$ of $AC$, intersecting $AD$ at $K$. If $\lvert KG \rvert = 3\text{ cm}$ and $\text{Area}(\triangle EFG) = 5\text{ cm}^2$, find the full median length $\lvert AD \rvert$ and the total area $\text{Area}(\triangle ABC)$.

Solution:

1. Apply the $3\text{-}1\text{-}2$ Rule for length: Along median $AD$, the segments satisfy $\lvert AK \rvert : \lvert KG \rvert : \lvert GD \rvert = 3k : k : 2k$. Since $\lvert KG \rvert = k = 3\text{ cm}$, the entire median is:

$$\lvert AD \rvert = 6k = 6 \times 3 = 18\text{ cm}$$

1. Area ratio via base and height scaling: First, midsegment $EF$ cuts off top triangle $\triangle AEF \sim \triangle ABC$ with similarity ratio $\frac{1}{2}$, so $\text{Area}(\triangle AEF) = \frac{1}{4}\text{Area}(\triangle ABC)$. Moreover, inside $\triangle AEF$ and $\triangle EFG$, both triangles share base $EF$, and their perpendicular heights are proportional to $\lvert AK \rvert = 3k$ and $\lvert KG \rvert = k$ (the shared sine of the angle between $AD$ and $EF$ cancels). Because areas with the same base are proportional to their altitudes:

$$\frac{\text{Area}(\triangle AEF)}{\text{Area}(\triangle EFG)} = \frac{3k}{k} = 3 \implies \text{Area}(\triangle AEF) = 3 \times 5 = 15\text{ cm}^2$$

Finally, $\text{Area}(\triangle ABC) = 4 \times \text{Area}(\triangle AEF) = 4 \times 15 = 60\text{ cm}^2$ (meaning $\text{Area}(\triangle EFG)$ is always $\frac{1}{12}$ of $\text{Area}(\triangle ABC)$!).

Review the Method

Triangles sharing a common vertex and altitude have areas directly proportional to their base segments (S₁ / S₂ = a₁ / a₂).