Advertisements
Advertisements
Question
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
Solution
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
RELATED QUESTIONS
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 `"QB"/"AQ" = "DR"/"AR"`
Rhombus PQRB is inscribed in ΔABC such that ∠B is one of its angle. P, Q and R lie on AB, AC and BC respectively. If AB = 12 cm and BC = 6 cm, find the sides PQ, RB of the rhombus.
DE || BC and CD || EE Prove that AD2 = AB × AF
Check whether AD is bisector of ∠A of ∆ABC of the following
AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm
∠QPR = 90°, PS is its bisector. If ST ⊥ PR, prove that ST × (PQ + PR) = PQ × PR
Construct a ∆PQR in which the base PQ = 4.5 cm, ∠R = 35° and the median from R to RG is 6 cm.
Construct a ∆ABC such that AB = 5.5 cm, ∠C = 25° and the altitude from C to AB is 4 cm
ABC is a triangle in which AB = AC. Points D and E are points on the side AB and AC respectively such that AD = AE. Show that the points B, C, E and D lie on a same circle
