ABC is a right triangle with angle B = 90°. A circle with BC as diameter meets hypotenuse AC at point D. Prove that :
(i) AC AD = (AB) ^2
(ii)(BD) ^2= AD x DC.
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Answers
Step-by-step explanation:
ΔABC is a right angled triangle.
∠ABC = 90°.
A circle is drawn with AB as diameter intersecting AC in P, PQ is the tangent to the circle which intersects BC at Q.
Join BP.
PQ and BQ are tangents drawn from an external point Q.
∴ PQ = BQ -------------- (1) (Length of tangents drawn from an external point to the circle are equal)
⇒ ∠PBQ = ∠BPQ (In a triangle, angles opposit to equal sides are equal)
Given that, AB is the diameter of the circle.
∴ ∠APB = 90° (Angle in a semi-circle is a right angle)
∠APB + ∠BPC = 180° (Linear pair)
∴ ∠BPC = 180° – ∠APB = 180° – 90° = 90°
Consider ΔBPC,
∠BPC + ∠PBC + ∠PCB = 180° (Angle sum property of a triangle)
∴ ∠PBC + ∠PCB = 180° – ∠BPC = 180° – 90° = 90° ----------- (2)
∠BPC = 90°
∴ ∠BPQ + ∠CPQ = 90° ...(3)
From equations (2) and (3), we get
∠PBC + ∠PCB = ∠BPQ + ∠CPQ
⇒ ∠PCQ = ∠CPQ (Since, ∠BPQ = ∠PBQ)
Consider ΔPQC,
∠PCQ = ∠CPQ
∴ PQ = QC ----------- (4)
From equations (1) and (4), we get
BQ = QC
Step-by-step explanation:
Step-by-step explanation:
ΔABC is a right angled triangle.
∠ABC = 90°.
A circle is drawn with AB as diameter intersecting AC in P, PQ is the tangent to the circle which intersects BC at Q.
Join BP.
PQ and BQ are tangents drawn from an external point Q.
∴ PQ = BQ -------------- (1) (Length of tangents drawn from an external point to the circle are equal)
⇒ ∠PBQ = ∠BPQ (In a triangle, angles opposit to equal sides are equal)
Given that, AB is the diameter of the circle.
∴ ∠APB = 90° (Angle in a semi-circle is a right angle)
∠APB + ∠BPC = 180° (Linear pair)
∴ ∠BPC = 180° – ∠APB = 180° – 90° = 90°
Consider ΔBPC,
∠BPC + ∠PBC + ∠PCB = 180° (Angle sum property of a triangle)
∴ ∠PBC + ∠PCB = 180° – ∠BPC = 180° – 90° = 90° ----------- (2)
∠BPC = 90°
∴ ∠BPQ + ∠CPQ = 90° ...(3)
From equations (2) and (3), we get
∠PBC + ∠PCB = ∠BPQ + ∠CPQ
⇒ ∠PCQ = ∠CPQ (Since, ∠BPQ = ∠PBQ)
Consider ΔPQC,
∠PCQ = ∠CPQ
∴ PQ = QC ----------- (4)
From equations (1) and (4), we get
BQ = QC