Given : ∠BAD=65∘,∠ABD=70∘,∠BDC =45∘
∵ABCD is a cyclic quadrilateral.
In △ABD,
∠BDA+∠DAB+∠ABD=180∘
(By using sum property of ∆0 )
∴‌‌∠BDA‌=180∘−(65∘+70∘)
‌=180∘−135∘=45∘
Now from △ACD,
∠ADC‌=∠ADB+∠BDC
‌=45∘+45∘
(∵∠BDA‌=∠ADB=45∘)
‌=90∘
Hence, ∠D makes right angle belongs in semi-circle therefore AC is a diameter of the circle.