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....

true

explanation:

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°