# 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.

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 XM$$^{2}$$ = YM ∙ MZ.

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, $$\frac{XM}{ZM}$$ = $$\frac{YM}{XM}$$

⟹ XM$$^{2}$$ = YM ∙ MZ. (Proved)

3. In the two similar triangles PQR and XYZ, PM ⊥ QR and XN ⊥ YZ. Prove that $$\frac{PQ}{XY}$$ = $$\frac{PM}{XN}$$.

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. $$\frac{PQ}{XY}$$ = $$\frac{PM}{XN}$$. (Proved) 3. Corresponding sides of similar triangles are proportional.