1.

In the given fig, AB is the diameter of the circle with centre O. The point C, D and E are on the circle such that AC = CD = DE = EB. Find `angleEBA`.

Answer» In the given figure, we can join `OC,OD and OE.`
Please refer to video to see the diagram.
As, `AC = CD = DE = EB`
`:. /_COA=/_DOC=/_EOD=/_BOE`
Let `/_COA=/_DOC=/_EOD=/_BOE = theta`
Then, `theta+theta+theta+theta = 180^@`
`=>theta = 45^@`
Now, in `Delta EOB`,
`OB = OE = ` Radius of the circle
`:. /_OEB = /_OBE`
Also,`/_OEB+/_OBE + /_BOE = 180^@`
`=>/_OBE+/_OBE +45^@ = 180^@`
`=>2/_OBE = 135^@`
`=>/_OBE = 67.5^@`
`:. /_EBA = /_OBE = 67.5^@`


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