Answer 1 :

Answer 2 :

Answer 3 :

In ΔDOC and ΔBOA,

AB || CD, thus alternate interior angles willbe equal,

∴∠CDO = ∠ABO

Similarly,∠DCO = ∠BAO

Also, for the two triangles ΔDOC and ΔBOA,vertically opposite angles will be equal;∴∠DOC = ∠BOA

Hence, by AAA similarity criterion,ΔDOC ~ ΔBOA

Thus, the corresponding sides areproportional.

DO/BO = OC/OA

⇒OA/OC = OB/OD

Hence, proved.

In the fig.6.36,QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.

Answer 4 :

Solution:

In ΔPQR,

∠PQR = ∠PRQ∴ PQ = PR ………………………(i)Given,

QR/QS = QT/PRUsing equation (i), we get

QR/QS = QT/QP……………….(ii)

In ΔPQS and ΔTQR, by equation (ii),

QR/QS = QT/QP∠Q = ∠Q∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]

Answer 5 :

In the figure, ifΔABE ≅ ΔACD, show thatΔADE ~ ΔABC.

Answer 6 :

In the figure,altitudes AD and CE of ΔABC intersect each other at the point P. Show that:

Answer 7 :

(i) ΔAEP ~ ΔCDP(ii) ΔABD ~ ΔCBE(iii) ΔAEP ~ ΔADB(iv) ΔPDC ~ ΔBEC

Answer 8 :

Given, E is a point on the side AD produced ofa parallelogram ABCD and BE intersects CD at F. Consider the figure below,.