and AE = . cm, show that DE BC Solution We have AB = . cm, AD = . cm, AC = .
cm and AE = . cm. BD AB AD = . – .
cm. AD DB = and AE EC = AD DB AE EC Therefore, by converse of Basic Proportionality Theorem , we have DE is parallel to BC . Hence proved. Fig.
. E Fig. . Geometry Example .
In the Fig. . , DE AC and DC AP . Prove that BE EC BC CP Solution In D BPA , we have DC AP .
By Basic Proportionality Theorem, we have BC CP BD DA … ( ) In D BCA , we have DE AC . By Basic Proportionality Theorem, we have, BE EC BD DA … ( ) From ( ) and ( ) we get, BE EC BC CP . Hence proved. Example .
In the Fig. . , AD is the bisector of Ð A . If BD = cm, DC = cm and AB = cm, find AC.
Solution In D ABC , AD is the bisector of Ð A By Angle Bisector Theorem BD DC AB AC = AC gives AC = . Hence, AC = = . cm Example . In the Fig.
. , AD is the bisector of Ð BAC , if AB = cm, AC = cm and BC = cm. Find BD and DC . Solution Let BD = x cm, then DC = ( – x )cm AD is the bisector of Ð A By Angle Bisector Theorem AB AC BD DC x gives x = we get, x = = .
cm Therefore, BD =