## In ABC, ray BD bisects ABC and ray CE bisects ACB. If seg AB seg AC then prove that ED || BC.

Given , In ABC ray BD bisects ABC .AB/BC = AD/CD ………..(i) [Angle bisector theorem]Since ray CE bisects ACBAC/BC = AE/BE …………….(ii) [Angle bisector theorem]Given seg AB = seg AC.Substitute AB in (ii)AB/BC = AE/BE…..(iii)From (i) AD/CD = AE/BE [in (i) AB/BC = AD/CD]EDBC [ converse of basic proportionality theorem]Hence proved.

## In the figure 1.44, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF.

In XDE, PQ || DE……. GivenXP/PD = XQ/QE…….. (I) (Basic proportionality theorem)In XEF, QR || EF……. GivenXR/RF = XQ/QE……. .(II) (Basic proportionality theorem)XP/PD = XR/RF ………. from (I) and (II)seg PR || seg DE ………. (converse of basic proportionality theorem)

## In ABC, seg BD bisects ABC. If AB = x, BC = x+5, AD = x-2, DC = x+2, then find the value of x.

Given BD bisects ABC.Also AB = x, BC = x+5AD = x-2, DC = x+2Since BD bisects ABC, AB/BC = AD/DC [Angle bisector theorem]x/(x+5) = (x-2)/(x+2)Cross multiplying, we getx(x+2) = (x+5)(x-2)x2+2x = x2+5x-2x-10×2+2x = x2+3x-10x = 10the value of x is 10 .

## In D LMN, ray MT bisects LMN. If LM = 6, MN = 10, TN = 8, then find LT.

Given ray MT bisects LMN.LM = 6MN = 10TN = 8Since ray MT bisects LMN , LM/MN = LT/TN [Angle bisector theorem]6/10 = LT/8LT = 6×8/10LT = 4.8Hence measure of LT is 4.8 units.

## In figure 1.41, if AB || CD || FE then find x and AE.

From figure BD = 8, DF = 4, AC = 12 and CE = xGiven ABCDFEBD/DF = AC/CE [Property of three parallel lines and their transversals]8/4 = 12/xx = 12×4/8x = 6CE = 6AE = AC+CEAE = 12+6AE = 18Hence measure of x is 6 units and AE is 18 units.

## Find QP using given information in the figure.

From figure MN = 25, NP = 40, MQ = 14Given NQ bisects MNP.MN/NP = MQ/QP [Angle bisector theorem]25/40 = 14/QPQP = 40×14/25QP = 22.5Hence measure of QP is 22.5 units.

## In trapezium ABCD, side AB ||side PQ ||side DC, AP = 15, PD = 12, QC = 14, find BQ.

Given ABPQDC.AP = 15PD = 12QC = 14AP/PD = BQ/QC [ Property of three parallel lines and their transversals]15/12 = BQ /14BQ = 15×14/12BQ = 17.5 units.Hence measure of BQ is 17.5 units.

## Measures of some angles in the figure are given. Prove that AP/PB = AQ/QC

ABC = 60˚ [Given]APQ = 60˚ [Given]Since the corresponding angles are equal, line PQBC.In ABC , PQBC.By basic proportionality theorem , AP/PB = AQ/QCHence proved.

## In MNP, NQ is a bisector of N. If MN = 5, PN = 7 MQ = 2.5 then find QP .

Given MN = 5, PN = 7, MQ = 2.5Since NQ is a bisector of N , PN/MN = QP/MQ [Angle bisector theorem]7/5 = QP/2.5QP = 7×2.5/5QP = 3.5Hence measure of QP is 3.5 units.

## In PQR, PM = 15, PQ = 25 PR = 20, NR = 8. State whether line NM is parallel to side RQ. Give reason.

Given PM = 15, PQ = 25, PR = 20, NR = 8PQ = PM+MQ25 = 15+MQMQ = 25-15MQ = 10PR = PN+NR20 = PN+8PN = 20-8PN = 12PM/MQ = 15/10 = 3/2PN/NR = 12/8 = 3/2In PQR , PM/MQ = PN/NR.By Converse of basic proportionality theorem , line NM side RQ.