Triangle Proofs

1. Given: $\triangle ABC$ with interior angles $\angle A$, $\angle B$, $\angle C$.

To prove: $\angle A + \angle B + \angle C = 180°$.

Proof:

1. Through vertex $A$, draw line $\ell$ parallel to $\overline{BC}$.

2. $\overline{AB}$ is a transversal cutting parallel lines $\ell$ and $\overleftrightarrow{BC}$. By the Alternate Interior Angles Theorem, the angle on the left side of $A$ along $\ell$ equals $\angle B$. Call this angle $\angle 1$, so $\angle 1 = \angle B$.

3. $\overline{AC}$ is a transversal cutting the same parallel lines. By the Alternate Interior Angles Theorem, the angle on the right side of $A$ along $\ell$ equals $\angle C$. Call this angle $\angle 2$, so $\angle 2 = \angle C$.

4. Angles $\angle 1$, $\angle A$, and $\angle 2$ together form a straight angle along line $\ell$ at point $A$. Therefore:

$\angle 1 + \angle A + \angle 2 = 180°$

5. Substituting from steps 2 and 3:

$\angle B + \angle A + \angle C = 180°$

$\blacksquare$

2. Given: $\triangle ABC$ with $AB = AC$. $M$ is the midpoint of $\overline{BC}$.

To prove: $\angle B \cong \angle C$.

Proof:

1. Draw segment $\overline{AM}$.

2. In $\triangle ABM$ and $\triangle ACM$:

   • $AB = AC$ (given),
   • $BM = CM$ ($M$ is the midpoint of $\overline{BC}$),
   • $AM = AM$ (reflexive property — shared side).

3. By the SSS (Side-Side-Side) congruence criterion, $\triangle ABM \cong \triangle ACM$.

4. Since corresponding parts of congruent triangles are congruent (CPCTC), $\angle B \cong \angle C$.

$\blacksquare$

3. (a) First find the third angles:

$\angle R = 180° - 47° - 65° = 68°$ $\angle Z = 180° - 47° - 65° = 68°$

All three pairs of corresponding angles are equal: $\angle P = \angle X = 47°$, $\angle Q = \angle Y = 65°$, $\angle R = \angle Z = 68°$.

By the AA (Angle-Angle) Similarity Postulate, $\triangle PQR \sim \triangle XYZ$.

(b) Since the triangles are similar, corresponding sides are proportional. Sides $PQ$ and $XY$ are corresponding (opposite the equal angles $\angle R$ and $\angle Z$):

$\frac{PQ}{XY} = \frac{QR}{YZ}$

$\frac{10}{15} = \frac{8}{YZ}$

$YZ = \frac{8 \times 15}{10} = 12 \text{ cm}$

4. (a) $D$ is the midpoint of $\overline{AB}$:

$D = \left(\frac{0+0}{2},\; \frac{6+0}{2}\right) = (0, 3)$

$E$ is the midpoint of $\overline{AC}$:

$E = \left(\frac{0+8}{2},\; \frac{6+0}{2}\right) = (4, 3)$

(b) Compute $DE$:

$DE = \sqrt{(4-0)^2 + (3-3)^2} = \sqrt{16} = 4$

Compute $BC$:

$BC = \sqrt{(8-0)^2 + (0-0)^2} = \sqrt{64} = 8$

Therefore $DE = 4 = \frac{1}{2} \cdot 8 = \frac{1}{2} \cdot BC$. $\checkmark$

For parallelism: $D(0,3)$ and $E(4,3)$ have the same $y$-coordinate, so $\overline{DE}$ is horizontal (slope $= 0$). $B(0,0)$ and $C(8,0)$ also have the same $y$-coordinate, so $\overline{BC}$ is horizontal (slope $= 0$).

Since both segments have slope $0$, $\overline{DE} \parallel \overline{BC}$. $\checkmark$

This confirms the Triangle Midsegment Theorem: the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length.

5. Finding $BD$:

In right triangle $\triangle ABC$ with right angle at $B$ and altitude $\overline{BD}$ to hypotenuse $\overline{AC}$, the altitude creates two smaller triangles. By the geometric mean relation:

$BD^2 = AD \cdot DC = 4 \times 9 = 36$

$BD = 6$

Proof of $BD^2 = AD \cdot DC$ in general:

Given: $\triangle ABC$ with $\angle B = 90°$. $D$ is the foot of the altitude from $B$ to $\overline{AC}$, so $\overline{BD} \perp \overline{AC}$.

To prove: $BD^2 = AD \cdot DC$.

1. In $\triangle ABD$: $\angle ADB = 90°$ (since $BD \perp AC$) and $\angle A = \angle A$ (shared angle).

2. In $\triangle ABC$: $\angle ABC = 90°$ (given) and $\angle A = \angle A$.

3. Therefore $\triangle ABD \sim \triangle ACB$ by AA similarity ($\angle A = \angle A$ and $\angle ADB = \angle ABC = 90°$, so $D \leftrightarrow B$ and $B \leftrightarrow C$).

4. Similarly, in $\triangle CBD$: $\angle BDC = 90°$ and $\angle C = \angle C$.

5. In $\triangle ABC$: $\angle ABC = 90°$ and $\angle C = \angle C$.

6. Therefore $\triangle CBD \sim \triangle CAB$ by AA similarity ($\angle C = \angle C$ and $\angle BDC = \angle ABC = 90°$, so $D \leftrightarrow A$ and $B \leftrightarrow B$).

7. Since $\triangle ABD \sim \triangle ACB$ and $\triangle CBD \sim \triangle CAB$, by transitivity $\triangle ABD \sim \triangle BCD$ (correspondence: $A \leftrightarrow B$, $B \leftrightarrow C$, $D \leftrightarrow D$).

8. From $\triangle ABD \sim \triangle BCD$, corresponding sides are proportional. Match the sides opposite the equal angles:

$\frac{AD}{BD} = \frac{BD}{DC}$

9. Cross-multiplying:

$BD^2 = AD \cdot DC$

$\blacksquare$