Question

1. Consider the isosceles triangle ABC, with AB = AC, and BAC = 20. Choose points E, D on the sides AB, AC, respectively, so that ZCBD = 60', and BCE = 50'. We will find LEDB. (i) Bring the parallel DF to BC, with F on AB. Connect points and F. and let K be the intersection of BD and CF. Show that DFBC is an isosceles trapezium. Mark all its angles. (ii) What type of triangles are BKC and FKD? What does that mean for BC and BK? (iii) What type of triangle is EBC? What does that mean for BC and BE? (iv) Based on (ii) and (iii), find the angle BEK. (v) What is the measure of the angle EKF? What type of triangle is FEK? (vi) What type of triangle is FEK? What does that mean for EF and EK? (vii) Compare triangles EFD and EDK. Find LEDB. А D E B

Answer 1

Giner fig A 20 AB - AC LBAC 20° LCBD = 60° LBCE - SO" Find LEDB ? FA In ABCG 160 40 480 60 LBGIC = 180- 150+60) K 50 110° 61 20 30 60 50 B (i) BC ll DF Rio 11 nuans I B C DF in ton isos cales traperium Accourding to parallel theorem <CBP = Z BDF = 60° (BD intersecting line) In A B C D <BDC = 180 - 160 +80 ) = -40° L BCF = L D FC = 60° (using parallel theoreen) In A B CF, <BFC = 180 - ( 80 +60) = 40° LBFD = LBFC & LDFC = 60+40=1000 IBA L. CDF - LBDF + LBDC = 100° LEB C = 80° LBC D = 80.

therefore, J B C DF in isos cales traperiun. 1) O BKC and OFFD are BC and BK both are equilamal triangle sides isosceles non equilateral triangle. equal. (10) DEBC BC and BE are the equal side. (iv) In A BEK LEBK - 20° LBK.E = 60° .ZBEK - 180 - (20+20) = 100° (W) LEKF A FEK (vi) &FEK .3 non equilatial toiangle non-equilamal triangle. un- equal sides are EF and Ek (vii's compare OEFD and OEDK Common edge ED LDKE = 1200 but LDFE = 1000