Which similarity statement is true? picture in the attached △cab∼△dba △bda∼△acb △cda∼△abc △adb∼△cda

Option 2. ΔBDA ∼ ΔACB is the right answer.

Step-by-step explanation:

As we know if we have two triangles similar we have to prove the following conditions.

1) All angles of these two angles should be similar.

2) Opposite sides of the angles should be in the same ratio.

So in the given picture we see the three triangles Δabd, Δabc and Δadc.

Now in Δabc and Δadc there is no similarity in angles or sides therefore these triangles are not similar.

Now we compare ΔBDA and ΔACB.

∠DAB =∠ACB = 90°and ∠ABD is a common angle shared by these two triangles.

We know that if two angles of the given triangles are same then the third angles will be same.

Therefore ∠ADB =∠CAB

Now we can say that all angles of ΔBDA and ΔACB are similar.

Hence ΔBDA ∼ ΔACB is the right answer.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

chase is right because he already been to the swings and he know how far it will be to get there

step-by-step explanation: