Home
Class 9
MATHS
[" A "AB" is a line seepment and "P" is ...

[" A "AB" is a line seepment and "P" is its mid-point.Dand "],[" Eare points on the same side of "AB" such that "],[/_BAD=/_ABE" and "/_EPA=/_DPB],[" (see "F" ."22" ).Show that "],[" (1) "ADAP" e "/_EBP],[" (i) "AD=BE]

Promotional Banner

Similar Questions

Explore conceptually related problems

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that angleBAD = angleABE and angleEPA = angleDPB (see figure). Show that AD=BE

AB is a line segment and P is its midpoint. D and E are points on the same side of AB such that angle BAD = angle ABE and angle EPA = angle DPB (see the given figure). Show that:

AB is a line segment and P is its midpoint. D and E are points on the same side of AB such that angleBAD = angleABE and angleEPA = angleDPB . Show that AD=BE.

AB is a line segment D and P is its mid-point. D and E are points on the same side of AB such that angleBAD = angleABE and angleEPA = angleDPB (see figure). Show that DeltaDAP~=DeltaEBP

AB is line segment and P is its mid point. D and E are points on the same side of AB such that BAD = ABE and EPA = DPB show that (i) Delta DAP ~= Delta EBP (ii) AD = BE

AB is a line segment and P is its midpoint. D and E are points on the same side of AD such that angleBAD=angleABE and angleEPA=angleDPB . Show that (i) DeltaDAP~=DeltaEBP, (ii) AD=BE.

AB is a line segment and P is its midpoint. D and E are points on the same side of AP such that angleBAD=angleABE and angleEPA=angleDPB . Show that (i) DeltaDAP~=DeltaEBP, (ii) AD=BE.

AB is a line and P is its midpoint. D and E are two points on the same side of line segment AB such that angleBAD=angleABE and angleEPA=angleDPB . Prove that AD=BE

D is a point on side BC of Delta ABC such that AD = AC (see Fig. 7.47). Show that AB > AD.

AB is a line - segment. P and Q are points on either side of AB such that each of them is equidistant from the points A and B (See Fig ). Show that the line PQ is the perpendicular bisector of AB.