Question

Дан треугольник ABC. AM и CH – высоты. Докажите подобие треугольников ABC и MBH. ​

Answer 1

Объяснение:

Дано: ΔАВС;

AM и CH – высоты.

Доказать:

ABC ~ MBH.

Доказательство:

1. Рассмотрим ΔАВМ и ΔНВС - прямоугольные.

∠В - общий.

• Два треугольника подобны, если два угла одного треугольника соответственно равны двум углам другого.

⇒ ΔАВМ ~ ΔНВС

Запишем отношение сходственных сторон:

$\displaystyle \frac{BH}{BM}=\frac{AB}{BC}$

2. Рассмотрим ΔABC и ΔMBH.

∠В - общий.

Из п.1 :

$\displaystyle \frac{BH}{BM} =\frac{AB}{BC}$   или   $\displaystyle \frac{BH}{AB}=\frac{BM}{BC}$

• Два треугольника подобны, если две стороны одного треугольника пропорциональны двум сторонам другого и углы, образованные этими сторонами, равны.

⇒ ΔABC ~ ΔMBH.

Answer 2

Proving the Similarity of Triangles ABC and MBH

To prove the similarity of triangles ABC and MBH, we can use the fact that if two triangles have two corresponding angles that are equal, then the triangles are similar. In this case, we can show that triangles ABC and MBH have two corresponding angles that are equal, which will prove their similarity.

First, let's consider triangle ABC and triangle MBH.

Step 1: Proving the Equality of Angles

1. Angle BAC is equal to angle BMH because both are right angles (as AM and CH are altitudes, and BMH is a right-angled triangle) [[1]]. 2. Angle ABC is equal to angle MBH because both are right angles (as AM and CH are altitudes, and ABC is a right-angled triangle) [[2]].

Step 2: Conclusion

Since we have shown that angle BAC is equal to angle BMH and angle ABC is equal to angle MBH, we can conclude that triangles ABC and MBH are similar by the Angle-Angle (AA) similarity criterion.

Therefore, we have successfully proved the similarity of triangles ABC and MBH.