Statmenet 1`:` The point of intesection of the common chords of three circles described on the three sides of a triangle as diameter is orthocentre of the triangle.
Statement-2 `:` The common chords of three circles taken two at a time are altitudes of the traingles.

Options

A. Statement-1 `:` is True, Statement-2 is True and Statement-2 is a correct explanation for Statement-1
B. Statement-1 is True, Statement-2 is True and Statement-2 is NOT a correct explanation for Statement -1
C. Statement-1 is True, Statement-2 is false
D. Statement-1 is False, Statement-2 is True

Answer

To solve the problem, we need to analyze the two statements given and prove their validity step by step.

### Step 1: Understanding the Statements
- **Statement 1**: The point of intersection of the common chords of three circles described on the three sides of a triangle as diameters is the orthocenter of the triangle.
- **Statement 2**: The common chords of the three circles taken two at a time are the altitudes of the triangle.

### Step 2: Define the Triangle and Circles
Let triangle ABC be given, where:
- Side AB is the diameter of circle C1.
- Side BC is the diameter of circle C2.
- Side AC is the diameter of circle C3.

### Step 3: Identify the Orthocenter
The orthocenter (O) of triangle ABC is the point where the altitudes of the triangle intersect. To find the orthocenter, we drop perpendiculars from each vertex to the opposite side:
- From vertex A to side BC (let's call this point D).
- From vertex B to side AC (let's call this point E).
- From vertex C to side AB (let's call this point F).

### Step 4: Analyze the Common Chords
When we consider the circles:
- The common chord of circles C1 and C2 is the line segment AD (altitude from A to BC).
- The common chord of circles C2 and C3 is the line segment BE (altitude from B to AC).
- The common chord of circles C3 and C1 is the line segment CF (altitude from C to AB).

### Step 5: Prove the Chords are Altitudes
Using the property of circles, we know that:
- If a line segment is a diameter of a circle, then any angle subtended by that diameter at any point on the circle is a right angle (90 degrees).
- Therefore, since AB is the diameter of circle C1, angle ADB is 90 degrees, making AD perpendicular to BC.
- Similarly, we can show that BE is perpendicular to AC and CF is perpendicular to AB.

### Step 6: Conclusion
From the analysis:
- The intersection point of the common chords (AD, BE, CF) is the orthocenter O of triangle ABC.
- The common chords are indeed the altitudes of the triangle.

Thus, both statements are true:
- **Statement 1** is true.
- **Statement 2** is true and provides a correct explanation for Statement 1.

### Final Answer
Both statements are true.

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