Question

In triangle ABC, Seg BD is perpendicular to side AC. Se DE is perpendicular to side BC then show that DE* BD = DC* BE

Answer 1

Answer:

The proof is explained below.

Step-by-step explanation:

Given triangle ABC, Seg BD is perpendicular to side AC. Seg DE is perpendicular to side BC we have to prove that $DE\times BD = DC\times BE$

Lets take the triangle BDE and triangle BDC

∠1 = ∠1            ( ∵ common)

∠3 = ∠4           ( ∵ each 90°)

Therefore, by AA similarity

ΔBDE is similar to ΔBDC

Therefore, $\frac{DE}{DC}=\frac{BD}{BC}=\frac{BE}{BD}$

⇒ $\frac{DE}{DC}=\frac{BE}{BD}$

⇒ $DE\times BD = DC\times BE$

Answer 2

Answer:

Step-by-step explanation:

In triangle ABC, segment BD is perpendicular to the side AC and segment DE is perpendicular to side BC.

In ΔBDC and ΔBDE, we have

∠1=∠2 (90°)

∠DBC=∠DBE(Common)

Therefore, by AA similarity,

ΔBDC ≅ΔBDE,

By similarity of triangles,

$\frac{BE}{BD}=\frac{DE}{DC}$

⇒$BE$×$DC$=$BD$×$DE$

Hence proved.