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Question 9

In Fig. 6.63, D is a point on side BC of Δ ABC such that $\dfrac{BD}{CD} = \dfrac{AB}{AC}$ Prove that AD is the bisector of $\angle BAC$.

Solution:

In the given figure, let us extend BA to P such that;

AP = AC.

Now join PC

Given, $\dfrac{BD}{CD} = \dfrac{AB}{AC}$

$\dfrac{BD}{CD} = \dfrac{AP}{AC}$

By using the converse of basic proportionality theorem, we get,

AD || PC

$\angle BAD = \angle APC$ (Corresponding angles) ……………….. (i)

And, $\angle DAC = \angle ACP$ (Alternate interior angles) …….… (ii)

By the new figure, we have;

AP = AC

$\angle APC = \angle ACP$ ……………………. (iii)

On comparing equations (i), (ii), and (iii), we get,

$\angle BAD = \angle APC$

Therefore, AD is the bisector of the angle BAC.

Hence, proved.

Additional Questions

Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Fig. 6.64)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Answer

In Fig. 6.56, PS is the bisector of $\angle QPR$ of Δ PQR. Prove that $\dfrac{QS}{SR} = \dfrac{PQ}{PR}$

Answer

In Fig. 6.57, D is a point on hypotenuse AC of Δ ABC, such that BD $\perp$ AC, DM $\perp$ BC and DN $\perp$ AB. Prove that :

(i) $DM^2 = DN . MC $

(ii) $DN^2 = DM . AN $

Answer

In Fig. 6.58, ABC is a triangle in which $\angle ABC > 90°$ and AD $\perp$ CB produced. Prove that $AC^2 = AB^2 + BC^2 + 2 BC . BD.$

Answer

In Fig. 6.59, ABC is a triangle in which $\angle ABC < 90°$ and AD $\perp$ BC. Prove that $AC^2 = AB^2 + BC^2 – 2 BC.BD.$

Answer

In Fig. 6.60, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :

(i)$ AC^2 = AD^2 + BC . DM + (\dfrac{BC}{2})^2$

(ii) $AB^2 = AD^2 – BC . DM + (\dfrac{BC}{2})^2$

(iii) $AC^2 + AB^2 = 2 AD^2 + \dfrac{1}{2} BC^2$

Answer

Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Answer

In Fig. 6.61, two chords AB and CD intersect each other at the point P. Prove that :

(i) Δ APC ~ Δ DPB

(ii) AP . PB = CP . DP

Answer

In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that

(i) Δ PAC ~ Δ PDB

(ii) PA . PB = PC . PD

Answer

In Fig. 6.63, D is a point on side BC of Δ ABC such that $\dfrac{BD}{CD} = \dfrac{AB}{AC}$ Prove that AD is the bisector of $\angle BAC$.

Answer
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