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

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## Thursday, July 29, 2010

### Exercise 10.5

Question 1:
In the given figure, A, B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC. Ans.

It can be observed that
∠AOC = ∠AOB + ∠BOC
= 60° + 30°
= 90°
We know that angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle. Question 2:
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Ans. In ΔOAB,
AB = OA = OB = radius
∴ ΔOAB is an equilateral triangle.
Therefore, each interior angle of this triangle will be of 60°.
∴ ∠AOB = 60° ⇒ ∠ADB = 180° − 30° = 150°
Therefore, angle subtended by this chord at a point on the major arc and the minor arc are 30° and 150° respectively.

Question 3:
In the given figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR. Ans. Consider PR as a chord of the circle.
Take any point S on the major arc of the circle.
∠PQR + ∠PSR = 180° (Opposite angles of a cyclic quadrilateral)
⇒ ∠PSR = 180° − 100° = 80°
We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∴ ∠POR = 2∠PSR = 2 (80°) = 160°
In ΔPOR,
OP = OR (Radii of the same circle)
∴ ∠OPR = ∠ORP (Angles opposite to equal sides of a triangle)
∠OPR + ∠ORP + ∠POR = 180° (Angle sum property of a triangle)
2 ∠OPR + 160° = 180°
2 ∠OPR = 180° − 160° = 20º
∠OPR = 10°

Question 4:
In the given figure, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC. Ans.

In ΔABC,
∠BAC + ∠ABC + ∠ACB = 180° (Angle sum property of a triangle)
⇒ ∠BAC + 69° + 31° = 180°
⇒ ∠BAC = 180° − 100º
⇒ ∠BAC = 80°
∠BDC = ∠BAC = 80° (Angles in the same segment of a circle are equal)

Question 5:
In the given figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC. Ans.

In ΔCDE,
∠CDE + ∠DCE = ∠CEB (Exterior angle)
⇒ ∠CDE + 20° = 130°
⇒ ∠CDE = 110°
However, ∠BAC = ∠CDE (Angles in the same segment of a circle)
⇒ ∠BAC = 110°

Question 6:
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.

Ans. For chord CD,
∠CBD = ∠CAD (Angles in the same segment)
∠BCD + 100° = 180°
∠BCD = 80°
In ΔABC,
AB = BC (Given)
∴ ∠BCA = ∠CAB (Angles opposite to equal sides of a triangle)
⇒ ∠BCA = 30°
We have, ∠BCD = 80°
⇒ ∠BCA + ∠ACD = 80°
30° + ∠ACD = 80°
⇒ ∠ACD = 50°
⇒ ∠ECD = 50°

Question 7:
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Ans. Let ABCD be a cyclic quadrilateral having diagonals BD and AC, intersecting each other at point O. (Consider BD as a chord)
∠BCD = 180° − 90° = 90° (Considering AC as a chord)
90° + ∠ABC = 180°
∠ABC = 90°
Each interior angle of a cyclic quadrilateral is of 90°. Hence, it is a rectangle.

Question 8:
If the non-parallel sides of a trapezium are equal, prove that it is cyclic

Ans.
. Consider a trapezium ABCD with AB | |CD and BC = AD.
Draw AM ⊥ CD and BN ⊥ CD.
In ΔAMD and ΔBNC,
∠AMD = ∠BNC (By construction, each is 90°)
AM = BM (Perpendicular distance between two parallel lines is same)
∴ ΔAMD ≅ ΔBNC (RHS congruence rule)
∴ ∠ADC = ∠BCD (CPCT) ... (1)
∠BAD + ∠BCD = 180° [Using equation (1)]
This equation shows that the opposite angles are supplementary.
Therefore, ABCD is a cyclic quadrilateral.

Question 9:
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD. Ans. Join chords AP and DQ.
For chord AP,
∠PBA = ∠ACP (Angles in the same segment) ... (1)
For chord DQ,
∠DBQ = ∠QCD (Angles in the same segment) ... (2)
ABD and PBQ are line segments intersecting at B.
∴ ∠PBA = ∠DBQ (Vertically opposite angles) ... (3)
From equations (1), (2), and (3), we obtain
∠ACP = ∠QCD

Question 10:
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Ans. Consider a ΔABC.
Two circles are drawn while taking AB and AC as the diameter.
Let they intersect each other at D and let D not lie on BC.
∠ADB = 90° (Angle subtended by semi-circle)
∠ADC = 90° (Angle subtended by semi-circle)
Therefore, BDC is a straight line and hence, our assumption was wrong.
Thus, Point D lies on third side BC of ΔABC. Question 11:
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.

Ans. In ΔABC,
∠ABC + ∠BCA + ∠CAB = 180° (Angle sum property of a triangle)
⇒ 90° + ∠BCA + ∠CAB = 180°
⇒ ∠BCA + ∠CAB = 90° ... (1)
∠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) • 5

sorry for that the answer is In ΔABC,
∠ABC + ∠BCA + ∠CAB = 180° (Angle sum property of a triangle)
⇒ 90° + ∠BCA + ∠CAB = 180°
⇒ ∠BCA + ∠CAB = 90° ... (1)
∠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 + ∠C
AB + ∠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)

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)

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)

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)

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)

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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