ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.

## Thursday, July 29, 2010

### Exercise 10.5

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sorry for that the answer is

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In ΔABC,

ABC + BCA + CAB = 180 (Angle sum property of a triangle)

⇒ 90 + BCA + CAB = 180

⇒ BCA + CAB = 90 ... (1)

In ΔADC,

CDA + ACD + DAC = 180 (Angle sum property of a triangle)

⇒ 90 + ACD + DAC = 180

⇒ ACD + DAC = 90 ... (2)

Adding equations (1) and (2), we obtain

BCA + CAB + ACD + DAC = 180

⇒ (BCA + ACD) + (CAB + DAC) = 180

BCD + DAB = 180 ... (3)

However, it is given that

B + D = 90 + 90 = 180 ... (4)

From equations (3) and (4), it can be observed that the sum of the measures of opposite angles of quadrilateral ABCD is 180. Therefore, it is a cyclic quadrilateral.

Consider chord CD.

CAD = CBD (Angles in the same segment)

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In ΔABC,

ABC + BCA + CAB = 180 (Angle sum property of a triangle)

⇒ 90 + BCA + CAB = 180

⇒ BCA + CAB = 90 ... (1)

In ΔADC,

CDA + ACD + DAC = 180 (Angle sum property of a triangle)

⇒ 90 + ACD + DAC = 180

⇒ ACD + DAC = 90 ... (2)

Adding equations (1) and (2), we obtain

BCA + CAB + ACD + DAC = 180

⇒ (BCA + ACD) + (CAB + DAC) = 180

BCD + DAB = 180 ... (3)

However, it is given that

B + D = 90 + 90 = 180 ... (4)

From equations (3) and (4), it can be observed that the sum of the measures of opposite angles of quadrilateral ABCD is 180. Therefore, it is a cyclic quadrilateral.

Consider chord CD.

CAD = CBD (Angles in the same segment)

- -1

In ΔABC,

ABC + BCA + CAB = 180 (Angle sum property of a triangle)

⇒ 90 + BCA + CAB = 180

⇒ BCA + CAB = 90 ... (1)

In ΔADC,

CDA + ACD + DAC = 180 (Angle sum property of a triangle)

⇒ 90 + ACD + DAC = 180

⇒ ACD + DAC = 90 ... (2)

Adding equations (1) and (2), we obtain

BCA + CAB + ACD + DAC = 180

⇒ (BCA + ACD) + (CAB + DAC) = 180

BCD + DAB = 180 ... (3)

However, it is given that

B + D = 90 + 90 = 180 ... (4)

From equations (3) and (4), it can be observed that the sum of the measures of opposite angles of quadrilateral ABCD is 180. Therefore, it is a cyclic quadrilateral.

Consider chord CD.

CAD = CBD (Angles in the same segment)

- -1

In ΔABC,

ABC + BCA + CAB = 180 (Angle sum property of a triangle)

⇒ 90 + BCA + CAB = 180

⇒ BCA + CAB = 90 ... (1)

In ΔADC,

CDA + ACD + DAC = 180 (Angle sum property of a triangle)

⇒ 90 + ACD + DAC = 180

⇒ ACD + DAC = 90 ... (2)

Adding equations (1) and (2), we obtain

BCA + CAB + ACD + DAC = 180

⇒ (BCA + ACD) + (CAB + DAC) = 180

BCD + DAB = 180 ... (3)

However, it is given that

B + D = 90 + 90 = 180 ... (4)

Consider chord CD.

CAD = CBD (Angles in the same segment)

- -4

In ΔABC,

ABC + BCA + CAB = 180 (Angle sum property of a triangle)

⇒ 90 + BCA + CAB = 180

⇒ BCA + CAB = 90 ... (1)

In ΔADC,

CDA + ACD + DAC = 180 (Angle sum property of a triangle)

⇒ 90 + ACD + DAC = 180

⇒ ACD + DAC = 90 ... (2)

Adding equations (1) and (2), we obtain

BCA + CAB + ACD + DAC = 180

⇒ (BCA + ACD) + (CAB + DAC) = 180

BCD + DAB = 180 ... (3)

However, it is given that

B + D = 90 + 90 = 180 ... (4)

Consider chord CD.

CAD = CBD (Angles in the same segment)

- -1