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Triangles

Long Answer Type

231.

In the given fig, PS is the bisector of ∠QPR of ∆PQR. Prove that $QS over SR equals PQ over PR.$

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232.

In the given fig., D is a point on hypotenuse AC of ∆ABC, DIM ⊥ BC and DN ⊥ AB.
Prove that:
(i) DM² = DN × MC.
(ii) DN² = DM × AN.

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Short Answer Type

233.

In the given fig., ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC² = AB² + BC² + 2 BC . BD.

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234.

In the given Fig, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC²= AB² + BC² - 2 BC.BD.

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Long Answer Type

235.

In the given fig, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
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(iii) $AC squared plus AB squared space equals space 2 AD squared plus 1 half BC squared.$

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

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237.

In the given fig, two chords AB and CD intersect each other at the point P. Prove that:
(i) ∆APC ~ ∆DPB.
(ii) AP . PB = CP . DP

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238.

In the given Fig, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that
(i) ∆PAC ~ ∆PDB.
(ii) PA. PB = PC . PD.

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239.
In the given fig, D is a point on side BC of ∆ABC such that $BD over CD equals AB over AC.$ Prove that AD is the bisector of ∠BAC.

Given: In figure, D is a point on side BC of AABC such that

BD over CD equals AB over AC.

To prove: AD is the bisector of ∠BAC i.e. ∠BAD = ∠CAD.
Construction : From BA produce cut off AE = A. Join CE.
Proof:

BD over CD equals AB over AC space space left square bracket Given right square bracket

rightwards double arrow

BD over CD equals AB over AE

[∵ AC = AE (by construction]
∴ In ∆BCE, we have
AD || CE

[By converse of Basic proportionality theorem]
∴ ∠1 = ∠3 ...(i) [Corres. ∠s]
and ∠2 = ∠4...(ii) [Alt. Int. ∠s]

∵ AC = AE [By construction]

∴ ∠3 = ∠4 ...(iii)

[Angles opposite equal sides of a triangle are equal]

Using (iii), (i) and (ii), we get

∠BAD = ∠CAD

Hence, AD is the bisector of ∠BAC.

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240. Nazima is fly fishing 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. 6.73)? 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 second ?

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