Q12 of 95 Page 6

In the given figure, ∠1 = ∠2 and , prove that ΔACB ~ ΔDCE

We have,


(∵, BD = DC as ∠1 = ∠2) …(i)


Also, ∠1 = ∠2


i.e. ∠DBC = ∠ACB


∴ ACB ~ DCE (by SAS similarity criterion)


Hence Proved


More from this chapter

All 95 →
10

In the given figure, if ∠A = ∠C, then prove that Δ AOB ~ Δ COD

 11

In the given figure DB ⊥ BC, DE ⊥ AB and AC ⊥ BC, prove that ΔBDE ~ΔABC.

 13

In an isosceles ΔABC with AC = BC, the base AB is produced both ways to P and Q such that AP x BQ = AC2. Prove that : ΔACP ~ ΔBQC

 14

In the given figure, find ∠P.