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)
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)
- -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)
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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