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
State, true or false:
The diagonals of a trapezium divide each other into proportional segments.
In the figure, given below, straight lines AB and CD intersect at P; and AC || BD. Prove that: If BD = 2.4 cm, AC = 3.6 cm, PD = 4.0 cm and PB = 3.2 cm; find the lengths of PA and PC.
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: ΔAOB is similar to ΔCOD.
In the given figure, AD = AE and AD2 = BD × EC. Prove that: triangles ABD and CAE are similar.
Triangle ABC is similar to triangle PQR. If bisector of angle BAC meets BC at point D and bisector of angle QPR meets QR at point M, prove that : `(AB)/(PQ) = (AD)/(PM)`.
In ΔABC, and E are the mid-points of AB and AC respectively. Find the ratio of the areas of ΔADE and ΔABC.
In the adjoining figure ABC is a right angle triangle with ∠BAC = 90°, and AD ⊥ BC.
(i) Prove ΔADB ∼ ΔCDA.
(ii) If BD = 18 cm, CD = 8 cm find AD.
(iii) Find the ratio of the area of ΔADB is to area of ΔCDA.
In the figure below, PB and QA are perpendiculars to the line segment AB. If PO = 6 cm, QO = 9 cm and the area of ΔPOB = 120 cm2, find the area of ΔQOA.
Triangles ABC and DEF are similar.
If area (ΔABC) = 9 cm2, area (ΔDEF) = 64 cm2 and BC = 5·1 cm find AB.
