Angle in a semicircle is a right angle.
AP.BP | = | (OP – OA) (OP – OB) |
= | ( r – a) ( r + a) | |
= | | r |2 – | a |2 | |
= | 0 (∵ | r | = | a | = radius) |
In any triangle, the altitudes are concurrent.
OA ⊥ BC | ⇒ | OA . BC = 0 |
⇒ | OA . (OC – OB) = 0 | |
⇒ | a . (c – b) = 0 | |
⇒ | a . c – a . b = 0 | |
⇒ | a . b = a . c ----- (i) | |
and OB ⊥ CA | ⇒ | OB . CA = 0 |
⇒ | OB . (OA – OC) = 0 | |
⇒ | b . (a – c) = 0 | |
⇒ | b . a – b . c = 0 (∵ b . a = a . b) | |
⇒ | a . b = b . c ------ (ii) | |
From (i) and (ii), a . c = b . c | ⇒ | c . b – c . a = 0 (∵ b . a = a . b) |
⇒ | c . (b – a) = 0 | |
⇒ | OC . (OB – OA) = 0 | |
⇒ | OC . AB = 0 |
In any triangle, the perpendicular bisectors of the sides are concurrent.
Now OD ⊥ BC | ⇒ | OD . BC = 0 |
⇒ | OD . (OC – OB) = 0 | |
⇒ | (c – b) = 0 | |
⇒ | (c2 – b2) = 0 ------ (i) | |
and OE ⊥ CA | ⇒ | OE . CA = 0 |
⇒ | OE . (OA – OC) = 0 | |
⇒ | (a – c) = 0 | |
⇒ | (a2 – c2) = 0 ----- (ii) | |
Adding (i) and (ii) we have | ⇒ | (a2 – b2) = 0 |
⇒ | (a – b) = 0 | |
⇒ | OF . (OA – OB) = 0 | |
⇒ | OF . BA = 0 | |
⇒ | OF⊥ BA |
In a parallelogram, the sum of the squares of the lengths of the diagonals is equals to sum of the squares of the lengths of its sides.
Now OB2 + CA2 | = | | a + c |2 + | a – c |2 |
= | (| a |2 + 2a . c + | c |2) + (| a |2 – 2a . c + | c |2) | |
= | 2 | a |2 + 2 | c |2 | |
= | | a |2 + | a |2 + | c |2 + | c |2 | |
= | OA2 + AB2 + CB2 + OC2 (∵ OA = BC; OC = AB) |
In ΔABC, the length of the median through the vertex A is .
4AD2 | = | α2 + β2 + 2 α. β |
= | AB2 + AC2 + 2 AB . AC | |
= | c2 + b2 + 2bc cos A | |
= | c2 + b2 + (b2 + c2 – a2) [∵ from cosine rule] | |
= | 2b2 + 2c2 – a2 | |
∴ AD | = |