Advertisements
Advertisements
प्रश्न
In ΔABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that: AB × BC = BP × CA
उत्तर
Consider ΔABC and ΔAPB
∠ABC = ∠APB ...[Exterior angle property]
∠BCP = ∠ABP ...[Given]
∴ ΔABC ∼ ΔAPB ...[AA criterion for similarity]
`(CA)/(AB) = (BC)/(BP)` ...(Corresponding sides of similar triangles are proportional)
`=>` AB × BC = BP × CA
APPEARS IN
संबंधित प्रश्न
In a trapezium ABCD, side AB is parallel to side DC; and the diagonals AC and BD intersect each other at point P. Prove that :
- ΔAPB is similar to ΔCPD.
- PA × PD = PB × PC.
P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, prove that: DP : PL = DC : BL.
P is a point on side BC of a parallelogram ABCD. If DP produced meets AB produced at point L, prove that: DL : DP = AL : DC.
In quadrilateral ABCD, the diagonals AC and BD intersect each other at point O. If AO = 2CO and BO = 2DO; show that: OA × OD = OB × OC.
In the given figure, AB || DC, BO = 6 cm and DQ = 8 cm; find: BP × DO.
In the following figure, ABCD to a trapezium with AB ‖ DC. If AB = 9 cm, DC = 18 cm, CF = 13.5 cm, AP = 6 cm and BE = 15 cm, Calculate: PE
In the following figure, AB, CD and EF are perpendicular to the straight line BDF.
If AB = x and CD = z unit and EF = y unit, prove that : `1/x + 1/y = 1/z`
ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that :
If AC = 13 cm, BC = 5 cm and AE = 4 cm. Find DE and AD.
In figure ABC and DBC are two triangles on the same base BC. Prove that
`"Area (ΔABC)"/"Area (ΔDBC)" = "AO"/"DO"`.
In the given figure ΔABC and ΔAMP are right angled at B and M respectively.
Given AC = 10 cm, AP = 15 cm and PM = 12 cm.
(i) Prove ΔABC ∼ Δ AMP.
(ii) Find AB and BC.
