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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.
On a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides, Prove that ∠BAC=∠BDC.
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Solution
Given ,
ACB and ADB are two right triangles
To Prove,
<BAC=<BDC
<Proof,
<C + <D = 90° + 90°
= 180°
Therefore ADBC is a Cyclic Quadrilateral
<BAC and <BDC are made by the same arc BC
Therefore,
<BAC=<BDC
Hence Proved
ACB and ADB are two right triangles
To Prove,
<BAC=<BDC
<Proof,
<C + <D = 90° + 90°
= 180°
Therefore ADBC is a Cyclic Quadrilateral
<BAC and <BDC are made by the same arc BC
Therefore,
<BAC=<BDC
Hence Proved
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