The side BC of a triangle ABC is bisected at D; O is any point i

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 Multiple Choice QuestionsLong Answer Type

291. Any point X inside ∆DEF is joined to its vertices. From a point P in DX, PQ is drawn parallel to DE meeting XE in Q and QR is drawn parallel to EF meeting XF in R. Prove that PR || DF.
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292. Let ABC be a triangle and D and E be two points on side AB such that AD = BE, then prove that PQ || AB.

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293. In a ∆ABC, D and E are points on sides AB and AC respectively such that BD = C’E. If ∠B = ∠C, show that DE || BC.

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294. Let X be any point on the side BC of a triangle ABC. If XM, XN are drawn parallel to BA and CA meeting CA, BA in M and N respectively, MN meets BC produced in T. Prove that TX2 = TB × TC.


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295. The side BC of a triangle ABC is bisected at D; O is any point in AD. BO and CO produced meet AC and AB in E and F respectively and AD is produced to X, So that D is the midpoint of OX. Prove that AO: AX = AF : AB and show that FE || BC.


Given: D is the mid point of BC as well as OX. CO and BO intersects each other at ‘O’ and their produced part meet AB and AC at F and E respectively.

Given: D is the mid point of BC as well as OX. CO and BO intersects e

To prove: (i) AO over AX equals AF over AB

(ii)    FE || BC
Const : Join B-X and C-X
Proof: We have,
BD = CD
and    OD = DX
⇒ BC and OX bisect each other.
⇒ OBXC is a parallelogram
⇒ BX || CO and CX || BO
⇒ BX || CF and CX || BE
⇒ BX || OF and CX || OE
Now in ∆ABX, we have
OF || BX
Therefore, By using Basic proportionality theorem, we have
AO over AX equals AF over AB         ...(i)
In ∆ACX, we have
OE || CX
Therefore, by using Basic proportionality theorem, we have
AO over AX equals AE over AC          ...(ii)
Comparing (i) and (ii), we get
AF over AB equals AE over AC
Therefore, by using converse of Basic proportionality theorem FE || BC.

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296. In the given figure: if AD over DC equals BE over EC and ∠CDE = ∠CED. Prove that ∆CAB is isosceles.

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297. P is the mid-point of side BC of ∆ABC. Q is the mid-point of AP. BQ when produced meets AC at L. Prove that <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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298. ABCD is a quadrilateral P, Q, R, S are the points of trisection of the sides AB, CB, CD and AD respectively. Prove that PQRS is a parallelogram.
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 Multiple Choice QuestionsShort Answer Type

299.

In the given Fig,  AB || DE and BC || EF. Prove that AC || DF.

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 Multiple Choice QuestionsLong Answer Type

300. Two triangles BAC and BDC, right angled at A and D respectively are drawn on the same base BC and on the same side of BC. If AC and DB intersect at E, prove that AE × EC = BE × DE. 

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