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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.

920 Views

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

Given: ∆CAB such that

AD over DC equals BE over EC

and

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To Prove: ∆CAB is an isosceles
Proof: It is given that

AD over DC equals BE over EC

rightwards double arrow

DC over AD equals EC over BC

(taking reciprocals of both sides)
Therefore by using converse of Basic proportionality theorem, we have
DE || AB
⇒    ∠CDE = ∠CAB [Corres. ∠s]
and    ∠CED = ∠CBA [Corres. ∠s]
But    ∠CDE = ∠CED    (given)
∴    ∠CAB = ∠CBA
⇒    BC = AC
[Sides opposite equal angles are equal]
Hence ∆CAB is an isosceles.

217 Views

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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370 Views

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.

1338 Views

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.

672 Views