Question

In the figure A,B,C,D,E are points on the circle

then prove that < A +< B+<C+<D+<E=180°

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Answer 1

We know that angles in the same segment of a circle are equal.

Thus, for chord CD, we have ∠1=∠2=∠3

For chord DE, we have ∠4=∠5=∠6

For chord EA, we have ∠7=∠8=∠9

For chord AB, we have ∠10=∠11=∠12

For chord BC, we have ∠13=∠14=∠15

Now, consider ∠DEB,∠EBC,∠BCD and ∠CDE.

Since EBCD is a quadrilateral, the sum of the angle is 360⁰

⇒m∠DEB+m∠EBC+m∠BCD+m∠CDE= 360⁰

⇒m∠3+m∠14+m∠5+m∠1+m∠10+m∠8+m∠6+m∠13+m∠11+m∠9=360⁰

⇒m∠A+m∠E+m∠B+m∠A+m∠D+m∠C+m∠B+m∠E+m∠D+m∠C=360⁰

⇒2(m∠A+m∠B+m∠C+m∠D+m∠E)=360⁰

⇒m∠A+m∠B+m∠C+m∠D+m∠E=180⁰.

Answer 2

For chord CD, we have ∠1= ∠2= ∠3

For chord DE, we have ∠4= ∠5= ∠6

For chord EA, we have ∠7= ∠8= ∠9

For chord AB, we have ∠10= ∠11= ∠12

For chord BC, we have ∠13= ∠14= ∠15

Now, consider ∠DEB, ∠EBC, ∠BCD and ∠CDE.

Since EBCD is a quadrilateral, the sum of the angle is 360⁰

⇒m∠DEB+ m∠EBC+ m∠BCD+ m∠CDE= 360⁰

⇒m∠3+ m∠14+ m∠5+ m∠1+ m∠10+ m∠8+ m∠6+ m∠13+ m∠11+ m∠9= 360⁰

⇒m∠A+ m∠E+ m∠B+ m∠A+ m∠D+ m∠C+ m∠B+ m∠E+ m∠D+ m∠C= 360⁰

⇒2(m∠A+ m∠B+ m∠C+ m∠D+ m∠E)= 360⁰

⇒m∠A+ m∠B+ m∠C+ m∠D+ m∠E= 180⁰