Question

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 $EB \cdot BD = AB \cdot BC$.

Answer 1

Answer:

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,

$\angle EAC = \angle ACD = 90^{\circ}$

To prove :

$EB\times BD=AB\times BC$

Proof :

In triangles AEC and EBA,

∠EAC= ∠ABE ( right angles )

∠CEA = ∠AEB ( common angles )

By AA similarity postulate,

$\triangle AEC \sim \triangle EBA$,

Similarly,

$\triangle AEC \sim \triangle ABC$

$\implies\triangle EBA\sim \triangle ABC-----(1)$

Now, In triangles ADC and CBD,

∠ACD = ∠CBD ( right angles )

∠ADC= ∠BDC ( common angles )

By AA similarity postulate,

$\triangle ADC \sim \triangle CBD$,

Similarly,

$\triangle ADC \sim \triangle ABC$

$\implies \triangle CBD\sim \triangle ABC-----(2)$

From equations (1) and (2),

$\triangle EBA\sim \triangle CBD$

The corresponding sides of similar triangles are in same proportion,

$\frac{EB}{BC}=\frac{AB}{BD}$

$\implies EB\times BD=AB\times BC$

Hence, proved....