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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.

We have,
AB ⊥ BC and DM ⊥ BC
⇒ AB || DM
Similarly, we have
CB ⊥ AB and DN ⊥ AB
⇒ CB || DN
Hence, quadrilateral BMDN is a rectangle.
∴ BM = ND
(i) In ∆BMD, we have
∠1 + ∠BMD + ∠2 = 180°
⇒ ∠1 + 90° + ∠2 = 180°
⇒ ∠1 + ∠2 = 90°
Similarly, in ∆DMC, we have
∠3 + ∠4 = 90°
Since BD ⊥ AC. Therefore
∠2 + ∠3 = 90°
Now, ∠1 + ∠2 = 90°and ∠2 + ∠3 = 90°
⇒ ∠1 + ∠2 = ∠2 + ∠3
⇒ ∠1 = ∠3
Also, ∠3 + ∠4 = 90° and ∠2 + ∠3 = 90°
⇒ ∠3 + ∠4 = ∠2 + ∠3 ⇒ ∠2 = ∠4
Thus, in ∆'s BMD and DMC, we have
∠1 = ∠3 and ∠2 = ∠4
Therefore, by using AA similar condition
 $increment BMD tilde increment DMC$

$space space space rightwards double arrow space space space space space space space space space space space space BM over DM equals MD over MC$

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$rightwards double arrow$ $DM squared space equals space DN cross times MC$

(ii) Proceeding as in (i), we can prove that
 $increment BND tilde increment DNA$
$rightwards double arrow space space space space space space space space space space space space space space space space space BN over DN equals ND over NA rightwards double arrow space space space space space space space space space space space space space space space space DM over DN space equals space DN over AN space left square bracket therefore space space space space BN space equals space DM right square bracket rightwards double arrow space space space space space space space space space space space space space space space space space space space DN squared space equals space DM space cross times space AN.$

706 Views

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.

386 Views

234.

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

181 Views

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) $AC squared plus AB squared space equals space 2 AD squared plus 1 half BC squared.$

81 Views

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.

1432 Views

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

2163 Views

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.

3086 Views

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.

240 Views

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 ?

2516 Views