In Fig. 6.39, ABC

(i)
 
To Prove: Δ ABC ∼ Δ AMP

Given: In ΔABC and ΔAMP,

∠ABC = ∠AMP (Each 90°)

Proof:

∠ABC = ∠AMP (Each 90°)

∠A = ∠A (Common)

∴ΔABC ΔAMP (By AA similarity)

Hence, Proved.

(ii) Δ ABC ∼ Δ AMP

Now we get that,

Similarity Theorem - If the lengths of the corresponding sides of two triangles are proportional, then the triangles must be similar. And the converse is also true, so we have


 

Hence, Proved.