Question

pls solve as fast as possible...(15pts)...

Answer 1

Hello Friend,

Here, we go step-wise as follows:

• O is centre of circle. Now line segment BC passes through O. This means that BC is a diameter of circle.

Also, O is the midpoint of BC.
So, OB = OC
Or, we can write:

BC = 2 OB = 2 OC ----(1)

• AB and AC are chords of a circle, and their end-points B and C form a diameter.

So, here, diameter BC subtends an angle CAB.

Now, any diameter always subtends a right angle on the circle.

So, CAB is a right angle.

So, by Pythagoras Theorem,

AC² + AB² = BC² ----(2)

• Now, OD is a line segment from centre O to the chord AB,

So, OD is a perpendicular bisector of AB.

AD = BD = ½ AB
Or, AB = 2 AD = 2 BD ---(3)

Also, ODB is a right angle.

• Now, ∆ODB is a right angled triangle.

So, By Pythagoras Theorem,

BD² + OD² = OB²
So, OD² = OB² - BD² ---(4)

• Now,
From (2)
AB² + AC² = BC²

So, (2 BD)² + AC² = (2 OB)² [From (1) and (3)]

So, 4 BD² + AC² = 4 OB²
So, AC² = 4 (OB² - BD²)
So, AC² = 4 (OD²) [From (4)]
So, AC² = 4 OD²
So, AC = 2 OD

Or, CA = 2 OD

Hence Proved.

__________________

There is an alternative approach, using similarity of triangles.

Here, in ∆CAB and ∆ODB,
ODB = CAB = 90°
AB : DB = 2 : 1
BC : OB = 2 : 1

So, by SAS (Side-Angle-Side), ∆ODB is similar to ∆CAB

Due to similarity, CA : OD = AB : DB
So, CA : OD = 2 : 1
So, CA = 2 OD

Hence Proved.


Hope it helps.

Purva
@Purvaparmar1405
Brainly.in

Answer 2

hi

O is centre of circle. Now line segment BC passes through O. This means that BC is a diameter of circle.

Also, O is the midpoint of BC.

So, OB = OC

Or, we can write:

BC = 2 OB = 2 OC ----(1)

• AB and AC are chords of a circle, and their end-points B and C form a diameter.

So, here, diameter BC subtends an angle CAB.

Now, any diameter always subtends a right angle on the circle.

So, CAB is a right angle.

So, by Pythagoras Theorem,

AC² + AB² = BC² ----(2)

• Now, OD is a line segment from centre O to the chord AB,

So, OD is a perpendicular bisector of AB.

AD = BD = ½ AB

Or, AB = 2 AD = 2 BD ---(3)

Also, ODB is a right angle.

• Now, ∆ODB is a right angled triangle.

So, By Pythagoras Theorem,

BD² + OD² = OB²

So, OD² = OB² - BD² ---(4)

• Now,

From (2)

AB² + AC² = BC²

So, (2 BD)² + AC² = (2 OB)² [From (1) and (3)]

So, 4 BD² + AC² = 4 OB²

So, AC² = 4 (OB² - BD²)

So, AC² = 4 (OD²) [From (4)]

So, AC² = 4 OD²

So, AC = 2 OD

Or, CA = 2 OD

Hence Proved.

__________________

There is an alternative approach, using similarity of triangles.

Here, in ∆CAB and ∆ODB,

ODB = CAB = 90°

AB : DB = 2 : 1

BC : OB = 2 : 1

So, by SAS (Side-Angle-Side), ∆ODB is similar to ∆CAB

Due to similarity, CA : OD = AB : DB

So, CA : OD = 2 : 1

So, CA = 2 OD

Hence Proved.