Question:
On a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides. Prove that ∠BAC = ∠BDC.
Solution:
Draw two right triangles ACB and ADB in a circle with centre O, where AB is the diameter of the circle.
Join CO.
We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.
Here, arc CB subtends ∠COB at the centre and ∠CAB at A on the circle.
∴ ∠COB = 2∠CAB ...(1)
Also, arc CB subtends ∠COB at the centre and ∠CDB at D on the circle.
∴ ∠COB = 2∠CDB ...(2)
Equating (1) and (2),
2∠CAB = 2∠CDB
⇒ ∠CAB = ∠CDB
Hence, ∠BAC = ∠BDC.