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Question
On a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides. Prove that ∠BAC = ∠BDC.
Solution
Given: ΔACB and ΔADB are two right angled triangles with common hypotenuse AB.
To prove: ∠BAC = ∠BDC
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 verticles.
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, ∠BAC and ∠BDC are angles of same segment BC.
∴ ∠BAC = ∠BDC
Hence proved.
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