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Triangles

Long 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.

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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

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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.

ABCD is a quadrilateral in which P, Q, R, S are the points of trisection of the sides AB, CB, CD and AD respectively.

To Prove: PQRS is a parallelogram.
Const : Join A-C
Proof: In ∆ABC, we have
ABCD is a quadrilateral in which P, Q, R, S are the points of trisect

Therefore, by using converse of Basic proportionality theorem, we have
PQ || AC ...(i)
In ∆ADC, we have
ABCD is a quadrilateral in which P, Q, R, S are the points of trisect

Therefore, by using converse of Basic proportionality theorem, we have
SR || AC ...(ii)
Comparing (i) and (ii), we get
SR || PQ
Similarly, we can prove
SP || RQ
Hence, PQRS is a parallelogram.

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Short Answer Type

299.

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

188 Views

Long 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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