Question

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Answer 1

Question

In the given figure, OB is perpendicular bisector of the lone segment DE, FA is perpendicular to OB and FE intersects OB at C. Prove that 1/OA + 1/OB = 2/OC

Solution

In ∆OAF and ∆OBD,

since AF is perpendicular to OB and DB is perpendicular to OB,

→ AF // BD (If corresponding angles are equal then lines are parallel)

→ angle OFA = angle ODB (corresponding angles)

Hence, by AA criterion of similarity, ∆OAF ~ ∆OBD

→ OA/OB = AF/BD .....(1)

In ∆FAC and ∆CBE

angle ECB = angle FCA (vertically opposite angles)

angle FAC = angle CBE (right angle each)

→ ∆FAC ~ ∆CBE (AA criterion of similarity)

→ AF/BE = AC/BC

Also, BE = BD

→ AF/BD = AC/BC .....(2)

Equating (1) and (2)

→ OA/OB = AC/BC

Here,

AC = OC - OA

BC = OB - OC

→ $\sf\frac{OA}{OB} = \frac{OC - OA}{OB - OC}$

Now, let OA = x, OB = y and OC = z

→ $\sf\frac{x}{y} = {z - x}{y - z}$

$\sf\frac{z - x}{y - z} - \frac{x}{y} = 0$

$\sf\frac{y(z - x) - x(y - z)}{y(y - z)} = 0$ (Taking LCM)

$\sf yz - xy - xy + xz = 0$ (cross multiplying)

→ yz + xz = 2xy

Now divide both sides by xyz

$\sf\frac{yz}{xyz} + \frac{xz}{xyz} = \frac{2xy}{xyz}$

$\sf\frac{1}{x} + \frac{1}{y} = \frac{2}{z}$

Now put x as OA, y as OB and z as OC

$\sf\frac{1}{OA} + \frac{1}{OB} = \frac{2}{OB}$

Hence Proved.