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 |
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2. ∆PQM ∼ ∆XYN |
2. By AA criterion of similarity. |
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3. \(\frac{PQ}{XY}\) = \(\frac{PM}{XN}\). (Proved) |
3. Corresponding sides of similar triangles are proportional. |
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