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 = 8x – 7, DB = 5x – 3, AE = 4x – 3 and EC = 3x – 1, find the value of x
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
In ΔABC, D and E are points on the sides AB and AC respectively. For the following case show that DE || BC
AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm.
If PQ || BC and PR || CD prove that `"QB"/"AQ" = "DR"/"AR"`
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
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 ∆PQR such that QR = 6.5 cm, ∠P = 60° and the altitude from P to QR is of length 4.5 cm
Draw a triangle ABC of base BC = 5.6 cm, ∠A = 40° and the bisector of ∠A meets BC at D such that CD = 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
ST || QR, PS = 2 cm and SQ = 3 cm. Then the ratio of the area of ∆PQR to the area of ∆PST is
