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Home / Geometry / Solution of exercise 8.3(Circle)| Class Nine – Ten (9 – 10)| Part – 2

# Solution of exercise 8.3(Circle)| Class Nine – Ten (9 – 10)| Part – 2

Problem – 4: The chords AB and CD of a circle with centre O meet at a right    angles at some points within the circle. Prove that ∠AOD +∠BOC=2 right angles.

General Enunciation: The chords AB and CD of a circle with centre O meet at right angles at some points within the circle. We have to prove that AOD+BOC=2 right angles.

Particular Enunciation: Given that the chords AB and CB of a circle with centre O meet at right angles at the point E within the circle. Let us join (O, A), (O, C), (O, B), and (O, D). Let us Prove that AOD+BOC= 2 right angles.

Construction: Let us join (B, D).

Proof: AOD and ABD are two angles at the centre and at the circumference respectively and on the same arc AD.

AOD=2ABD ———————–(1)

Similarly, BOC and BDC are two angles at the centre and at the circumference respectively and on the same arc BC.

BOC=2BDC ———————— (2)

By adding equations (1) and (2) we get

AOD+BOC=2(ABD+BDC) ——————- (3)

But in right triangle BED, BED=1 right angle

EBD+BDE=1 right angle

Or,ABD+BDC= 1right angle ———————–(4)

From equation (3) and (4) we get,                                                                                                                  AOD+BOC=2(1 right angle)=2 right angles).  (Proved)

Problem – 5: If the vertical angles of two triangles standing on equal bases are supplementary, prove that their circum-circles are equal.

General Enunciation: If the vertical angles of two triangles standing on equal bases are supplementary, we have to prove that their circum-circles are equal.

Particular Enunciation: Suppose, in two triangles ABC and DEF, base BC =base EF and vertical angles are A and D, where A+D=2 right angles let us prove that circum-circles of ABC and DEF are equal.

ProofIn the circle ABHC with centre at P, BPC and BAC are two angles at the centre and at the circumference respectively and one the same are BHC.

BPC =2BAC =2 A ———————– (1)

In the circle DERF with centre at Q, EQF and EDF are two angles at the centre and at the circumference respectively and one the same are ERF.

EQF =2EDF =D ————————–(2)

By adding equations (i) and (ii) we get

BPC =EQF= 2 (A+ D)

=2×2.right angles

=4 right angles.

Since the sum of the angles at centre is 4 right angles and since BC=EF then are subtended by BC and EF are equal.

i.e., Minor are BAC = Minor are ERF

And Major are BHC = Major are EDF

∴ Arc BAC+ arc BHC =arc ERF +arc EDF

∴ The circum-circle of ABC= circum-circle of DEF (proved)

Problem – 6: The opposite angles of the quadrilateral ABCD are supplementary to each other. If the line AC is the bisector of ∠ BAD, prove that BC=CD.

General Enunciation: The opposite angles of the quadrilateral ABCD   are supplementary to each other. If the line AC is the bisector of BAD,   we have to prove that BC=CD.

Particular Enunciation: Given that opposite angles of the quadrilateral ABCD are supplementary to each other and the line AC is the bisector   of BAD .Let us prove that BC =CD

Proof: Since the opposite angles of the quadrilateral ABCD are supplementary to each other so the quadrilateral ABCD is cyclic.

AC is the bisector of BAD

Then we get, CAD and CAB are two angles at the circumference one arc CD and BC respectively

.but since CAD = CAB,

∴ Arc BC =arc CD

∴ BC=CD (Proved).

# For 1st part : CLICK On Part -1

## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.

If two triangles have the three sides of the one equal to the three sides ...