Question

6. In Fig. 13.26, the congruent parts are marked. Prove that ∆BAC = ∆CDB ​

Answer 1

Answer:

To prove :- ∆ BAC ≈ ∆ CDB

Given that <A = 90°, <D= 90°, AB=DC

In ∆ BAC and ∆ CDB

BC = BC (Common)

<A = <D = 90°

AB = DC (Given)

so ∆BAC ≈ ∆ CDB by SAS rule

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