Advertisements
Advertisements
Question
In ΔABC, D and E are points on the sides AB and AC respectively. For the following case show that DE || BC
AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm
Advertisements
Solution
Here AB = 12 cm, BD =12 – 8 = 4 cm, AE =12 cm, EC = 18 – 12 = 6 cm
∴ `"AD"/"DB" = 8/4` = 2
`"AE"/"EC" = 12/6` = 2
`"AD"/"DB" = "AE"/"EC"`
By converse of basic proportionality theorem DE || BC
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
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 `"AR"/"AD" = "AQ"/"AB"`
If PQ || BC and PR || CD prove that `"QB"/"AQ" = "DR"/"AR"`
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"`
∠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.
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
ST || QR, PS = 2 cm and SQ = 3 cm. Then the ratio of the area of ∆PQR to the area of ∆PST is
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.
