Exercise 49 Page 767

Statements	Reasons
1. MHT is a semicircle, RH⊥TM	1. Given
2. ∠ THM is a right angle	2. If an inscribed angle intercepts a semicircle, the angle is a right angle
3. ∠ TRH is a right angle	3. Definition of perpendicular lines
4. ∠ THM ≅ ∠ TRH	4. All right angles are congruent
5. ∠ T ≅ ∠ T	5. Reflexive Property of Congruence
6. ∠ TRH ~ ∠ THM	6. AA Similarity Theorem
7. TR/TH = RH/HM	7. Definition of similar polygons
8. TR/RH = TH/HM	8. Rearranging the equation

Practice makes perfect

We are given a circle centered at point R and △ TMH inscribed in the circle such that RH⊥TM. Then, we have that ∠ TRH is a right angle.

Since MHT is a semicircle, its measure is 180^(∘). Then, the measure of the inscribed angle ∠ THM is 90^(∘). Therefore, ∠ THM ≅ ∠ TRH and also ∠ T ≅ ∠ T by the Reflexive Property of Congruence.

Applying the Angle-Angle (AA) Similarity Theorem, we get that △ TRH ~ △ THM. Finally, by the definition of similar polygons, the lengths of corresponding sides are proportional. TR/TH = RH/HM ⇒ TR/RH = TH/HM

Given: & MHT is a semicircle, RH⊥TM Prove: & TRRH = THHM Let's summarize the proof we did above in the following two-column proof table.

Statements	Reasons
1. MHT is a semicircle, RH⊥TM	1. Given
2. ∠ THM is a right angle	2. If an inscribed angle intercepts a semicircle, the angle is a right angle
3. ∠ TRH is a right angle	3. Definition of perpendicular lines
4. ∠ THM ≅ ∠ TRH	4. All right angles are congruent
5. ∠ T ≅ ∠ T	5. Reflexive Property of Congruence
6. ∠ TRH ~ ∠ THM	6. AA Similarity Theorem
7. TR/TH = RH/HM	7. Definition of similar polygons
8. TR/RH = TH/HM	8. Rearranging the equation

Hints