In a triangle BD is perpendicular to AC . Prove that AB²-BC²+AC²= 2AC.AD
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GIVEN: In triangle ABC :
angle B =90°; BD perpendicular AC
TO PROVE: AB²- BC²+ AC² = 2*AC*AD
PROOF: In triangle ABC and triangle ADB
Angle A = Angle A (common)
Angle B = Angle D (90°)( BD perpendicular AC)
=> Triangle ABC ~ Triangle ADB (A.A)
=> AB = BC = AC (EQ...1)
AD DB AB
Similarly, Triangle ABC ~ Triangle BDC
=> AB = BC = AC (EQ...2)
BD DC BC
From eq. 1 we have AB = AC
AD AB
=> AB² = AC* AD
Similarly from eq. 2 we have
BC = AC
DC BC
=> BC²= AC* DC
WE KNOW THAT AC² = AC*AC
= AC( AD + DC) ( FROM FIGURE)
THUS;
AB²-BC² + AC² =(AC*AD) - (AC*DC ) +( AC(AD +DC)
= AC*AD - AC*DC + AC*AD + AC*DC
= AC*AD + AC*AD
= 2*AC*AD
Hope it helps....
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