Question

in the figure seg BD is perpendicular to side AC seg DE is perpendicular to side BC then show that into DE ×BD=DC×BE

Answer 1

Answer:  

Given : $BD\perp AC$  and $DE\perp BC$

To Prove : $DE \times BD = DC\times BE$

Proof:

Join the point B and C   (construction)

In triangles BDC and BED,

$\angle BDC\cong \angle BED$ ( Right angles)

$\angle DBC\cong \angle DBC$

By AA similarity postulate,

$\triangle BDC\cong \triangle BED$

By the property of similar triangles,

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

⇒ $DE \times BD = DC\times BE$

Hence proved.

Answer 2

Answer:

Given : BD\perp ACBD⊥AC and DE\perp BCDE⊥BC

To Prove : DE \times BD = DC\times BEDE×BD=DC×BE

Proof:

Join the point B and C (construction)

In triangles BDC and BED,

\angle BDC\cong \angle BED∠BDC≅∠BED ( Right angles)

\angle DBC\cong \angle DBC∠DBC≅∠DBC

By AA similarity postulate,

\triangle BDC\cong \triangle BED△BDC≅△BED

By the property of similar triangles,

\frac{BE}{BD} = \frac{DE}{CD}
BD
BE
​
=
CD
DE
​


⇒ DE \times BD = DC\times BEDE×BD=DC×BE

Hence proved.