Q. 4

# 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.

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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