Advertisements
Advertisements
प्रश्न
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.
In a quadrilateral ABCD, shown in the following figure, AB = AD. The bisectors of ∠BAC and ∠CAD meet the sides BC and CD at the points E and F, respectively. Prove that EF || BD.
Advertisements
उत्तर
ABCD is a quadrilateral, AB = AD.
AE and AF are the internal bisectors of ∠BAC and ∠DAC
.
To prove: EF || BD.
Construction: Join EF and BD
Proof: In ∆ ABC, AE is the internal bisector of ∠BAC.
By the angle bisector theorem, we have,
∴ `"AB"/"AC" = "BE"/"EC"` ...(1)
In ∆ ADC, AF is the internal bisector of ∠DAC.
By the angle bisector theorem, we have,
`"AD"/"AC"= "DF"/"FC"`
∴ `"AB"/"AC" = "DF"/"FC"` ...(AB = AD given) ...(2)
From (1) and (2), we get,
`"BE"/"EC" = "DF"/"FC"`
Hence, in ∆ BCD,
BD || EF ...(by converse of BPT)
Hence proved.
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
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.
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"`
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
Construct a ∆PQR in which the base PQ = 4.5 cm, ∠R = 35° and the median from R to RG is 6 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
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.
