On a common hypotenuse AB two right angled triangles ACB and ADB are drawn such that they lie on the opposite sides. Prove that ∠BAC = ∠BDC.

Given ACB and ADB are two right angled triangles having common hypotenuse AB.
We have to prove that ∠BAC = ∠BDC.
Construction: Join CD.
Proof:
⇒∠C + ∠D = 90° + 90° = 180°
We know that if opposite angles of a quadrilateral are supplementary, then it is a cyclic quadrilateral.
∴ ADBC is a cyclic quadrilateral.
We know that if two angles are of the same arc, then they are equal.
Here, ∠BAC and ∠BDC are made by the same arc BC.
∴ ∠BAC = ∠BDC
Hence proved
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