There are two problems based on similarity of triangles in RD Sharma for 10th
1. D is the mid point of side BC of a triangle ABC.AD is bisected at point E and BE produced cuts AC at X. Prove that BE:EX=3:1.
2.ABCD is parallelogram and APQ is a straight line meeting BC at P and DC produced at Q.Prove that the rectangle obtained by BP and DQ is equal to rectangle obtained by AB & BC.
NO FIGURES HAVE BEEN PROVIDED FOR BOTH.
Oct 2nd 2009, 07:39 AM
1) rotate the triangle AEX around the point E by 180 degrees and you'll obtain triangle DEX', as on the picture. Now DX' and CX are parallel (because DX' and AX are) and D is the midpoint of BC, so X' is the midpoint of BX (a basic result you should understand). Now since the length of XE is 1/2 of the length of XX' which is 1/2 of the lenght of XB, we have the length of XE is 1/4 of the length of XB.
As for 2) it would be polite to define what "two rectangles are equal" mean. If they mean that the rectangles have equal area, then since triangles ABP and QCP are similar, we have
BP/AB = CP/CQ
CQ.BP = AB.CP
AB.BP + CQ.BP = AB.BP + AB.CP
(AB+CQ).BP = AB.(BP+CP)
DQ.BP = AB.BC