Question

in the figure seg BD is perpendicular to side ac. seg de is perpendicular to side bc then show that de x bd= dc x be

Answer 1

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}BDBE​=CDDE​

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

Hence proved.