if ABCD is a rhombus diagonal AC and BD intersect at O and E is point lying on the circle with center O then the sum of angle BAE and EDC
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AC is diameter of circle.
So , a(ABC) = 90°
But ∆ABC is an isosceles ∆ with AB = BC So, a(BAC) = a(BCA) = 90/2 = 45°
a(BAC) = a(BDC) = 45°
Also , a(BAE) = a(BDE)
a(BAE) + a(EDC) = a(BDE) + a(EDC) =
a(BDC) = 45°
So , a(ABC) = 90°
But ∆ABC is an isosceles ∆ with AB = BC So, a(BAC) = a(BCA) = 90/2 = 45°
a(BAC) = a(BDC) = 45°
Also , a(BAE) = a(BDE)
a(BAE) + a(EDC) = a(BDE) + a(EDC) =
a(BDC) = 45°
Harshhp:
how a(BAE) = (BDE) = 45'
Angles in the same segment are equal
how it is 45'
I have to write 45° in the above line..
ooooooo so soory bro
my apologies
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