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:

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:

9th Grade Math

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