In figure 1.75, A – D – C and B – E – C seg DE || side AB If AD = 5, DC = 3, BC = 6.4 then find BE.

In figure 1.74, PM = 10 cm A(Δ PQS) = 100 sq.cm A (ΔQRS) = 110 sq.cm then find NR.

ΔMNT ~ ΔQRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio

In the figure 1.76, seg PA, seg QB, seg RC and seg SD are perpendicular to line AD.

AB = 60, BC = 70, CD = 80, PS = 280 then find PQ, QR and RS.

In ΔPQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR.

Complete the proof by filling in the boxes. In ΔPMQ, ray MX is bisector of ∠PMQ.

.......... (I) theorem of angle bisector.

In ΔPMR, ray MY is bisector of ∠PMR.

.......... (II) theorem of angle bisector.

But .......... M is the midpoint QR, hence MQ = MR.

∴ XY || QR .......... converse of basic proportionality theorem.