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AA Criterion of Similarity

Here we will prove the theorems related to AA Criterion of Similarity on Quadrilateral.

1. 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.

Solution:

Given: Let XYZ be a right angle in which ∠YXZ = 90° and XM ⊥ YZ.

AA Criterion of Similarity

Therefore, ∠XMY = ∠XMZ = 90°.

To prove: ∆XYM ∼ ∆ZXM ∼ ∆ ZYX.

Proof:

            Statement

            Reason

1. In ∆XYM and ∆XYZ,

(i) ∠XMY = ∠YXZ = 90°.

(ii) ∠XYM = ∠XMZ

1.

(i) Given.

(ii) Common angle.

2. Therefore, ∆XYM ∼ ∆ZYX.

2. By AA criterion of similarity.

3. In ∆XYZ and ∆XMZ,

(i) ∠YXZ = ∠XMZ = 90°.

(ii) ) ∠XZY= ∠XZM.

3.

(i) Given.

(ii) Common angle.

4. Therefore, ∆ZYX ∼ ∆ ZXM.

4. By AA criterion of similarity.

5. Therefore, ∆XYM ∼ ∆ZXM ∼ ∆ ZYX. (Proved)

5. From statement 2 and 4.


2. If in the ∆XYZ, ∠X = 90° and XM ⊥ YZ, M being the foot of the perpendicular, prove that XM2 = YM ∙ MZ.

AA Criterion of Similarity Problem

Solution:

In ∆XMY and ∆ZMX,

∠XMY = ∠ZMX = 90°

∠YXM = ∠XZM, because ∠XYM + ∠YXM = 90° = ∠XZM + ∠XYM

⟹ ∠YXM = ∠XZM

Therefore, ∆XMY ∼ ∆ZMX, (by AA criterion of similarity)

Therefore, XMZM = YMXM

⟹ XM2 = YM ∙ MZ. (Proved)


3. In the two similar triangles PQR and XYZ, PM ⊥ QR and XN ⊥ YZ. Prove that PQXY = PMXN.

Solution:

Proof:

            Statement

            Reason

1. In ∆PQM and ∆XYN,

(i) ∠PQM = ∠XYN

(ii) ∠PMQ = ∠XNY = 90°

1.

(i) Being similar triangles, they are equiangular.

(ii) Given

2. ∆PQM ∼ ∆XYN

2. By AA criterion of similarity.

3. PQXY = PMXN. (Proved)

3. Corresponding sides of similar triangles are proportional.








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