**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=2**∠**ABD ———————–(1)

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

∴ **∠**BOC=2**∠**BDC ———————— (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.

**Proof: **In 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 =2**∠**BAC =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 =2**∠**EDF =**∠**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 PDF: CLICK ON** ———– Solution of exercise 8.3.docx _Class 9 – 10_

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