AA Criterion of Similarity
Here we will prove that in a right-angled triangle, if a perpendicular is drawn from the right-angled vertex to the hypotenuse, the triangles on each side of it are similar to the whole triangle and to one another.
Given: Let XYZ be a right angle in which ∠YXZ = 90° and XM ⊥ YZ.
Therefore, ∠XMY = ∠XMZ = 90°.
To prove: ∆XYM ∼ ∆ZXM ∼ ∆ ZYX.
Proof:
|
Statement 1. In ∆XYM and ∆XYZ, (i) ∠XMY = ∠YXZ = 90°. (ii) ∠XYM = ∠XMZ 2. Therefore, ∆XYM ∼ ∆ZYX. 3. In ∆XYZ and ∆XMZ, (i) ∠YXZ = ∠XMZ = 90°. (ii) ) ∠XZY= ∠XZM. 4. Therefore, ∆ZYX ∼ ∆ ZXM. 5. Therefore, ∆XYM ∼ ∆ZXM ∼ ∆ ZYX. (Proved) |
Reason 1. (i) Given. (ii) Common angle. 2. By AA criterion of similarity. 3. (i) Given. (ii) Common angle. 4. By AA criterion of similarity. 5. From statement 2 and 4. |
From AA Criterion of Similarity to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
|
|
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.