
Vector Geometry: We can study different properties and relations relating to geometry with the help of vectors. Such a study is known as the Vector Geometry.
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Here are 12 geometrical theorems and
their proofs by vector method.
Theorem 1: Midpoint Formula

Theorem 2: Section Formula for Internal Division

Theorem 3: Section Formula: External Division

Theorem 4: Centroid Formula

Theorem 5:

Theorem 6:
![Theorem 6: Prove by vector method that the line joining vertex and midpoint of the base of an isosceles triangle is perpendicular to the base. Proof: Let ABC be an isosceles triangle where AB = AC and AM is the median to the base BC. Let ("AB" ) ā = ( "a" ) ā and ("AC" ) ā = ( "b" ) ā. Then, |( "a" ) ā| = |( "b" ) ā| i.e. a = b. And, ("BM" ) ā = "1" /"2" ("BC" ) ā By triangle law of vector addition, ("BC" ) ā = ("BA" ) ā + ("AC" ) ā = ā ("AB" ) ā + ("AC" ) ā = ā (" a " ) ā + (" b " ) ā = ((" b " ) ā ā (" a " ) ā) ("AM" ) ā = ("AB" ) ā + ("BM" ) ā = ("AB" ) ā + "1" /"2" ("BC" ) ā = (" a " ) ā + "1" /"2" (ā (" a " ) ā + (" b " ) ā) = ("2" ( "a" ) ā - ( "a" ) ā + ( "b" ) ā)/"2" = (( "a" ) ā + ( "b" ) ā)/"2" = "1" /"2" ((" a " ) ā" +" (" b " ) ā ) Now, ("AM" ) ā . ("BC" ) ā = "1" /"2" ((" a " ) ā" +" (" b " ) ā ) . ((" b " ) ā ā (" a " ) ā) = "1" /"2" (b2 ā a2) = 0 [āµ a = b] ā“ AM ā„ BC proved.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiLm_HIZzIWPeZE9FxRm0gWYhTKq99q7FQvwYTnCKE8UKwiWFdmBU7yqVV-AnHaea0cBN9deVyDRKRn_rZGAd3yc0U0ZVEQHQUFSRedjgcurGBV3NvIW0RLN1FJRta2hOTkby8Fv3go8768/s16000/theorem+6.png)
Theorem 7:

Theorem 8:

Theorem 9:

Theorem 10:
![Theorem 10: Prove by vector method that the diagonals of a rhombus bisect each other at right angles. Proof: Let OABC is a rhombus, ("OA" ) ā = ("CB" ) ā = (" a " ) ā and ("OC" ) ā = ("AB" ) ā = (" b " ) ā where |(" a " ) ā| = |(" b " ) ā| i.e. a = b. Let M be the midpoint of diagonal OB and N is the midpoint of diagonal AC. Now, by triangle law of vector addition, ("OB" ) ā = ("OA" ) ā + ("AB" ) ā = (" a " ) ā + (" b " ) ā. Since, M is the midpoint of OB, ("OM" ) ā = "1" /"2" ("OB" ) ā = "1" /"2" ((" a " ) ā + (" b " ) ā) ā¦ā¦ā¦ā¦.. (i) Again, since N is the midpoint of AC, by midpoint theorem, ("ON" ) ā = "1" /"2" (("OA" ) ā + ("OC" ) ā) = "1" /"2" ((" a " ) ā + (" b " ) ā) ā¦ā¦ā¦ā¦.. (ii) From (i) and (ii), the midpoints M and N of diagonals OB and AC coincide each other, i.e. they are the same point. Hence, the diagonals of a rhombus bisect each other. Now, ("AC" ) ā . ("OB" ) ā = (ā( "a" ) ā + ( "b" ) ā ) . (( "a" ) ā + ( "b" ) ā) = ((" b " ) ā)2 ā ((" a " ) ā)2 = b2 ā a2 = 0 [āµ a = b] ā“ ("AC" ) ā ā„ ("OB" ) ā. Hence diagonals of a rhombus bisect each other at right angle. Proved.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgs9FJlJlaqmLPk6Ct5V5GEF_83EhZIH5pGWAKarpe-OKuljwYb3qA8aJysiLSidFElq3rZGyScjHhcB2lY-A2kpl1icukHUvhEKE4YYVBkj6TNeIIYZtKRN63mxs4eJ2elx3qnj4F_5nnS/s16000/theorem+10.png)
Theorem 11:
![Theorem 11: Prove by vector method that the angle in the semi-circle is a right angle. Proof: Let O be the centre of a circle and AB be a diameter. ā“ ("AO" ) ā = ("OB" ) ā. ā ACB is an angle in the semi-circle. Join OC. AO = OB = OC (Radii of same circle). By triangle law of vector addition, ("AC" ) ā = ("AO" ) ā + ("OC" ) ā ("CB" ) ā = ("CO" ) ā + ("OB" ) ā = ("CO" ) ā + ("AO" ) ā = ("AO" ) ā ā ("OC" ) ā Now, ("AC" ) ā . ("CB" ) ā = ((AO) ā+(OC) ā ) . (("AO" ) ā ā ("OC" ) ā) = (("AO" ) ā)2 ā (("OC" ) ā)2 = AO2 ā OC2 = 0 [āµ AO = OC] ā“ ā ACB = 90°. Proved.](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjIxQib3jHHm9Sx-gTmVgiK-fNAnzSJnssnMo1XtmiwA1SqAZxFwJvJe9tEVWFkk1Qxc0HOFOS1B3KDCpf8cCecQiWlCIfj2RdJUaDqRPO_y7yrllFuHUa2J1Fb0mhVGtx135VN7VQaCT3H/s16000/theorem+11.png)
Theorem 12:

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