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