Triangle $abc$ has a right angle at $b$. legs $\overline{ab}$ and $\overline{cb}$ are extended past point $b$ to points $d$ and $e$, respectively, such that $\angle eac = \angle acd = 90^\circ$. prove that $eb \cdot bd = ab \cdot bc$.

a,d,e

Step-by-step explanation:

Given :

ABC is a right triangle in which ∠ABC = 90°,

Also, Legs AB and CB are extended past point B to points D and E,

Such that,

To prove :

Proof :

In triangles AEC and EBA,

∠EAC= ∠ABE ( right angles )

∠CEA = ∠AEB ( common angles )

By AA similarity postulate,

,

Similarly,

Now, In triangles ADC and CBD,

∠ACD = ∠CBD ( right angles )

∠ADC= ∠BDC ( common angles )

By AA similarity postulate,

,

Similarly,

From equations (1) and (2),

The corresponding sides of similar triangles are in same proportion,

Hence, proved....

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.

answer: i'm great at algebra

~coco

step-by-step explanation: