Question

In triangle ABC, angle BAC=90, AD is the bisector of angle BAC and DE is perpendicular to AC. Prove that DEx(AB+AC)=ABxAC.

Answer 1

Answer:

DEx(AB+AC)=ABxAC.

Step-by-step explanation:

In triangle ABC, angle BAC=90, AD is the bisector of angle BAC and DE is perpendicular to AC. Prove that DEx(AB+AC)=ABxAC.

Let draw DF ⊥ AB

Then ΔADE & ΔADF

AD is  Common

∠E = ∠F = 90°

∠DAF = ∠DAE ( ∠A bisector)

=> ΔADE & ΔADF are corrugent

=> DE = DF

Area of ΔABD  & ΔACD

= (1/2)AB * DF  & (1/2)AC * DE

DF = DE

=> (1/2)AB * DE  & (1/2)AC * DE

Area  of Δ ABC = Area of ΔABD  + Area of ΔACD

=> Area  of Δ ABC =  (1/2)AB * DE  + (1/2)AC * DE

=>  Area  of Δ ABC = (1/2)DE ( AB + AC)

Area of Δ ABC = (1/2) AB * AC

=> (1/2)DE ( AB + AC) = (1/2) AB * AC

=> DEx(AB+AC)=ABxAC.

QED

Proved