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
Answers
Answered by
5
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.
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.
Similar questions