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On a common hypotenuse AB, two right ang...

On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. Prove that `angleBAC=angleBDC`.

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Given, `DeltaACB and Delta ADB` are two right angled triangles with common hypotenuse AB.
To prove `angleBAC=angleBDC`
Construction join CD.
Proof Let O be the mid-point of AB.
Then, OA=OB =OC=OD.
Since, mid-point of the hypotenuse of a right triangle is equidistant from its vertices. Now, draw a circle to pass through the points A, B, C and D. with O as centre and radius equal to OA.
We know that, angles in the same segment of a circle are equal. From the figure, `angleBAC and angleBDC` are angles of same segment BC.
`:. angleBAC= angleBDC`
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