Advertisements
Advertisements
प्रश्न
If PQ || BC and PR || CD prove that `"QB"/"AQ" = "DR"/"AR"`

Advertisements
उत्तर
In ∆ABC, PQ || BC ...(Given)
By basic proportionality theorem
`"AP"/"PC" = "AQ"/"QB"` ...(1)
In ∆ADC, PR || CD ...(Given)
By basic proportionality theorem
`"AP"/"PC" = "AR"/"RD"` ...(2)
From (1) and (2) we get
`"AQ"/"QB" = "AP"/"RD"`
or
`"QB"/"AQ" = "DR"/"AR"`
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
If PQ || BC and PR || CD prove that `"AR"/"AD" = "AQ"/"AB"`

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"`
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
ABCD is a quadrilateral in which AB = AD, the bisector of ∠BAC and ∠CAD intersect the sides BC and CD at the points E and F, respectively. Prove that EF || BD.
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 ∆PQR such that QR = 6.5 cm, ∠P = 60° and the altitude from P to QR is of length 4.5 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.
