Question

Is △DBE similar to △ABC ? If so, which postulate or theorem proves these two triangles are similar?

Answer 1

DBE is similar to ABC by the SAS Similarity Theorem

Answer 2

Answer:

Step-by-step explanation:

In the given figure we have two triangles (One into another).

In triangle BDE,

$DB=10\ cm$

$BE=16\ cm$

In triangle ABC,

$AB=BD+AD=10+15=25\ cm$

$BC=CE+EB=24+16=40\ cm$

Now, in ΔABC and ΔBDE , we have

$\angle{B}=\angle {B}$                [Reflexive property]

$\dfrac{BD}{AB}=\dfrac{10}{25}=\dfrac{2}{5}=\dfrac{16}{40}=\dfrac{BE}{CE}$

By SAS Similarity Theorem ​,

ΔDBE ​ is similar to ​ ΔABC ​

• SAS Similarity Theorem say that if two sides in a triangle are proportional to two sides in another triangle and the included angle in both are congruent then the two triangles are said to be  similar.