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प्रश्न
In the given figure, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25°.
Find:
- ∠CAD
- ∠CBD
- ∠ADC
उत्तर
In the given figure,
ABCD is a cyclic quad in which AB || DC
∴ ABCD is an isosceles trapezium
∴ AD = BC
i. Join BD
Ext. ∠BCE = ∠BAD ...[Exterior angle of a cyclic qud is equal to interior opposite angle]
∴ ∠BAD = 80° ...[∵ ∠BCE = 80°]
But ∠BAC = 25°
∴ ∠CAD = ∠BAD – ∠BAC
= 80° – 25°
= 55°
ii. ∠CBD = ∠CAD ...[Angle of the same segment]
= 55°
iii. ∠ADC = ∠BCD ...[Angles of the isosceles trapezium]
= 180° – ∠BCE
= 180° – 80°
= 100°
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