Question

in triangle abc ad perpendicular BC such that ad square is equal to bd.CD prove that triangle ABC is right angle at a

Answer 1

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I may be wrong in this, still giving it a try:

Given Conditions- AD _|_ BC & AD^2= BD X DC

We have to prove Angle BAC= 90 Degree

If Angle BAC= 90 Degree

Then, BC^2= AB^2 + AC^2——————-(1)

As per the figure (BC= BD + DC)

Then (1) becomes,

(BD+DC)^2= AB^2 + AC^2

BD^2 + DC^2 + 2.BD.DC= AB^2 + AC^2

we have AD^2= BD.DC

BD^2 + DC^2 + (2.AD^2)= AB^2 + AC^2———(2)

As per figure AD^2= AC^2 - DC^2

Hence (2) becomes,

BD^2+ DC^2+2.AC^2–2.DC^2= AB^2+AC^2

We get

BD^2 + AC^2= AB^2 + DC^2

replacing AC^2 and AB^2 through Pythagoras theorem;

BD^2 + AD^2 +DC^2= AD^2 + BD^2+DC^2

which gives us LHS=RHS

Hence BC^2 = AB^2 + AC^2 which proves that BC is hypotenuse and Angle BAC is 90 Degree.

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