ABC ànd ADC are 2 right ∆s with common hypotenuseAC. Prove angle CAD=CBD
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Since, ABC and ADC are 2 right angle triangles with common hypotenuse AC i.e. angle B and angle D = 90
Hence, we can draw a circle passing through ABCD with AC as the diameter. (The angle subtended by the diameter of a circle on the circumference is 90)
Hence, now, DC is the chord of the circle and angle CAD and angle CBD are the angles subtended by the chord on the circumference of the circle.
Since, angles subtended by the same chord on the major arc of the circle are equal. So, angle CAD = angle CBD.
Hence proved.
Hence, we can draw a circle passing through ABCD with AC as the diameter. (The angle subtended by the diameter of a circle on the circumference is 90)
Hence, now, DC is the chord of the circle and angle CAD and angle CBD are the angles subtended by the chord on the circumference of the circle.
Since, angles subtended by the same chord on the major arc of the circle are equal. So, angle CAD = angle CBD.
Hence proved.
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Since, ABC and ADC are 2 right angle triangles with common hypotenuse AC then the right angles are also equal.
If we draw a circle passing through ABCD with AC as the diameter. then it is ABCD will be cyclic
So DC will be the chord of the circle and angle CAD and angle CBD will be the angles subtended by the chord on the circumference of the circle.
From the theorem of angles in the same segment are equal it is proved that angle CAD is equal to angle CBD
If we draw a circle passing through ABCD with AC as the diameter. then it is ABCD will be cyclic
So DC will be the chord of the circle and angle CAD and angle CBD will be the angles subtended by the chord on the circumference of the circle.
From the theorem of angles in the same segment are equal it is proved that angle CAD is equal to angle CBD
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