Advertisements
Advertisements
प्रश्न
In trapezium ABCD, AB || DC, E and F are points on non-parallel sides AD and BC respectively, such that EF || AB. Show that = `"AE"/"ED" = "BF"/"FC"`
Advertisements
उत्तर
Given: ABCD is a trapezium AB || DC
E and F are the points on the side of AD and BC
EF || AB
To Prove: `"AE"/"ED" = "BF"/"FC"`
Construction: Join AC intersecting AC at P
Proof:
In ∆ABC, PF || AB ...(Given)
By basic proportionality theorem
`"AP"/"PC" = "BF"/"FC"` ...(1)
In the ∆ACD, PE || CD ...(Given)
By basic Proportionality theorem
`"AP"/"PC" = "AE"/"ED"` ...(2)
From (1) and (2) we get
`"AE"/"ED" = "BF"/"FC"`
APPEARS IN
संबंधित प्रश्न
In ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC
If `"AD"/"DB" = 3/4` and AC = 15 cm find AE
ABCD is a trapezium in which AB || DC and P, Q are points on AD and BC respectively, such that PQ || DC if PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD
If PQ || BC and PR || CD prove that `"AR"/"AD" = "AQ"/"AB"`
If PQ || BC and PR || CD prove that `"QB"/"AQ" = "DR"/"AR"`
DE || BC and CD || EE Prove that AD2 = AB × AF
∠QPR = 90°, PS is its bisector. If ST ⊥ PR, prove that ST × (PQ + PR) = PQ × PR
Construct a ∆PQR in which QR = 5 cm, ∠P = 40° and the median PG from P to QR is 4.4 cm. Find the length of the altitude from P to QR.
Construct a ∆ABC such that AB = 5.5 cm, ∠C = 25° and the altitude from C to AB is 4 cm
Draw ∆PQR such that PQ = 6.8 cm, vertical angle is 50° and the bisector of the vertical angle meets the base at D where PD = 5.2 cm
Two circles intersect at A and B. From a point, P on one of the circles lines PAC and PBD are drawn intersecting the second circle at C and D. Prove that CD is parallel to the tangent at P.
