ABC is a right triangle. Angle C is 90 degree.
Altitude CD is drawn to hypotenuse AB.
Prove AB/BC = BC/BD and AB/AC = AC/AD
This is a theorem. How do I prove it? Thanks.
Dont you have similar triangles? You can show this by looking at the angles involved...remember the altitude forms a perpendicular angle on the hypotenuse.
ABC WITH A=35,AC=6cm and C=65 a second triangle wit G=80,GH=9cm and H=35
To prove the theorem in question, we can start by considering the similar triangles formed by the altitudes of the right triangle ABC.
Let's break down the proof step by step:
Proof for AB/BC = BC/BD:
1. Considering triangle ABC, we have two similar triangles formed - triangle ABC and triangle CDB.
2. Notice that both triangles have a right angle (angle C) and angle B is common to both.
3. By the angle-angle similarity theorem, we can conclude that triangles ABC and CDB are similar.
4. Since these triangles are similar, the corresponding sides are in proportion.
Therefore, we have AB/BC = BC/BD.
Proof for AB/AC = AC/AD:
1. Similarly, considering triangle ABC, we have two similar triangles formed - triangle ABC and triangle ADC.
2. In this case, both triangles have a right angle (angle C) and angle A is common to both.
3. By the angle-angle similarity theorem, we can conclude that triangles ABC and ADC are similar.
4. Therefore, the corresponding sides are in proportion, resulting in AB/AC = AC/AD.
In summary, by establishing the similarity between the triangles formed by the altitudes, we can prove that AB/BC = BC/BD and AB/AC = AC/AD using the angle-angle similarity theorem.