Q4 of 63 Page 150

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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