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**Q.1. In the figure, PS is the bisector of âˆ QPR of Î” PQR. Prove that **

**Sol.**

Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.

Given that, PS is the angle bisector of âˆ QPR.

âˆ QPS = âˆ SPR â€¦ (1)

By construction,

âˆ SPR = âˆ PRT (As PS || TR) â€¦ (2)

âˆ QPS = âˆ QTR (As PS || TR) â€¦ (3)

Using these equations, we obtain

âˆ PRT = âˆ QTR

âˆ´ PT = PR

By construction,

PS || TR

By using basic proportionality theorem for Î”QTR**Q.2. In the figure, D is a point on hypotenuse AC of Î”ABC, such that BD âŠ¥ AC, DM âŠ¥ BC and DN âŠ¥ AB. **

(i) Let us join DB.

We have, DN || CB, DM || AB, and âˆ B = 90Â°

âˆ´ DMBN is a rectangle.

âˆ´ DN = MB and DM = NB

The condition to be proved is the case when D is the foot of the perpendicular drawn from B to AC.

âˆ´ âˆ CDB = 90Â°

â‡’ âˆ 2 + âˆ 3 = 90Â° â€¦ (1)

In Î”CDM,

âˆ 1 + âˆ 2 + âˆ DMC = 180Â°

â‡’ âˆ 1 + âˆ 2 = 90Â° â€¦ (2)

In Î”DMB,

âˆ 3 + âˆ DMB + âˆ 4 = 180Â°

â‡’ âˆ 3 + âˆ 4 = 90Â° â€¦ (3)

From equation (1) and (2), we obtain

âˆ 1 = âˆ 3

From equation (1) and (3), we obtain

âˆ 2 = âˆ 4

In Î”DCM and Î”BDM,

âˆ 1 = âˆ 3 (Proved above)

âˆ 2 = âˆ 4 (Proved above)

âˆ´ Î”DCM âˆ¼ Î”BDM (AA similarity criterion)

â‡’ DM^{2} = DN Ã— MC

(ii) In right triangle DBN,

âˆ 5 + âˆ 7 = 90Â° â€¦ (4)

In right triangle DAN,

âˆ 6 + âˆ 8 = 90Â° â€¦ (5)

D is the foot of the perpendicular drawn from B to AC.

âˆ´ âˆ ADB = 90Â°

â‡’ âˆ 5 + âˆ 6 = 90Â° â€¦ (6)

From equation (4) and (6), we obtain

âˆ 6 = âˆ 7

From equation (5) and (6), we obtain

âˆ 8 = âˆ 5

In Î”DNA and Î”BND,

âˆ 6 = âˆ 7 (Proved above)

âˆ 8 = âˆ 5 (Proved above)

âˆ´ Î”DNA âˆ¼ Î”BND (AA similarity criterion)

â‡’ DN^{2} = AN Ã— NB

â‡’ DN^{2} = AN Ã— DM (As NB = DM)**Q.3. In the figure, ABC is a triangle in which âˆ ABC >90Â° and AD âŠ¥ CB produced. Prove that AC ^{2} = AB^{2} + BC^{2} + 2 BC.BD.**

AB

Applying Pythagoras theorem in Î”ACD, we obtain

AC

AC

AC

AC

Thus, we have:

AC

AD

â‡’ AD

Applying Pythagoras theorem in Î”ADC, we obtain

AD

AB

AB

AC

= AB

AM

Applying Pythagoras theorem in Î”AMC, we obtain

AM

AM

(AM

AD

Using the result, DC = BC/2 , we obtain

(ii) Applying Pythagoras theorem in Î”ABM, we obtain

AB

= (AD

= (AD

= AD

= AD

(iii) Applying Pythagoras theorem in Î”ABM, we obtain

AM

Applying Pythagoras theorem in Î”AMC, we obtain

AM

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

2AM

2AM

2AM

2AM

Let ABCD be a parallelogram.

Let us draw perpendicular DE on extended side AB, and AF on side DC.

Applying Pythagoras theorem in Î”DEA, we obtain

DE

Applying Pythagoras theorem in Î”DEB, we obtain

DE

DE

(DE

DA

Applying Pythagoras theorem in Î”ADF, we obtain

AD

Applying Pythagoras theorem in Î”AFC, we obtain

AC

= AF

= AF

= (AF

AC

Since ABCD is a parallelogram,

AB = CD â€¦ (iv)

And, BC = AD â€¦ (v)

In Î”DEA and Î”ADF,

âˆ DEA = âˆ AFD (Both 90Â°)

âˆ EAD = âˆ ADF (EA || DF)

AD = AD (Common)

âˆ´ Î”EAD Î”FDA (AAS congruence criterion)

â‡’ EA = DF â€¦ (vi)

Adding equations (i) and (iii), we obtain

DA

DA

BC

[Using equations (iv) and (vi)]

AB

(i) Î” APC ~ Î” DPB

(i) In Î”APC and Î”DPB,

âˆ APC = âˆ DPB (Vertically opposite angles)

âˆ CAP = âˆ BDP (Angles in the same segment for chord CB)

Î”APC âˆ¼ Î”DPB (By AA similarity criterion)

(ii) We have already proved that

Î”APC âˆ¼ Î”DPB

We know that the corresponding sides of similar triangles are proportional.

(i) In Î”PAC and Î”PDB,

âˆ P = âˆ P (Common)

âˆ PAC = âˆ PDB (Exterior angle of a cyclic quadrilateral is âˆ PCA = âˆ PBD equal to the opposite interior angle)

âˆ´ Î”PAC âˆ¼ Î”PDB

(ii)We know that the corresponding sides of similar triangles are proportional.

âˆ´ PA.PB = PC.PD

It is given that,

By using the converse of basic proportionality theorem, we obtain

AD || PC

â‡’ âˆ BAD = âˆ APC (Corresponding angles) â€¦ (1)

And, âˆ DAC = âˆ ACP (Alternate interior angles) â€¦ (2)

By construction, we have

AP = AC

â‡’ âˆ APC = âˆ ACP â€¦ (3)

On comparing equations (1), (2), and (3), we obtain

âˆ BAD = âˆ APC

â‡’ AD is the bisector of the angle BAC.

Let AB be the height of the tip of the fishing rod from the water surface. Let BC be the horizontal distance of the fly from the tip of the fishing rod.

Then, AC is the length of the string.

AC can be found by applying Pythagoras theorem in Î”ABC.

AC

AB

AB

AB

Thus, the length of the string out is 3 m.

She pulls the string at the rate of 5 cm per second.

Therefore, string pulled in 12 seconds = 12 Ã— 5 = 60 cm = 0.6 m

Let the fly be at point D after 12 seconds.

Length of string out after 12 seconds is AD.

AD = AC âˆ’ String pulled by Nazima in 12 seconds

= (3.00 âˆ’ 0.6) m

= 2.4 m

In Î”ADB,

AB

(1.8 m)

BD

BD = 1.587 m

Horizontal distance of fly = BD + 1.2 m

= (1.587 + 1.2) m

= 2.787 m

= 2.79 m

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