Question

prove it pls guys,, ,,,,,,...

Answer 1

first see traingle ACB which is right angle traingle .
so, use Pythagoras theorem .
AB^2 = AC^2 +BC^2 --------------(1)

again triangle BCP also right angle traingle .
so,
BP^2 =CP^2 +BC^2 ------------------(2)

subtracting equation (1) and (2)
AB^2 -BP^2 =AC^2 -CP^2
AB^2 =AC^2 +BP^2-CP^2
=AC^2 +BP^2-(AC-AP)^2
=AC^2+BP^2-AC^2-AP^2+2AC.AP
=BP^2 - AP^2+2AC.AP--------(3)

we also know ,
by circle property
AP.CP =BP.DP-----(4){see text book

now ,
equation (3)
AB^2 =BP^2 -AP^2+ AC.AP+AC.AP
=BP^2-AP(AP-AC)+AC.AP
=BP^2 -AP.CP +AC.AP
but from equation (4)
AP.CP =BP.DP
put this
AB^2 = BP^2 -BP.DP+AC.AP

AB^2=BP (BP-DP) +AC.AP

AB^2=BP.BD +AC.AP
hence proved

Answer 2

In triangle ACB (right angled triangle) .
AB^2 = AC^2 +BC^2 --------------(1)

In triangle BCP (right angled triangle) . 
BP^2 =CP^2 +BC^2 --------------(2)

From (1) and (2)
AB^2 -BP^2 =AC^2 -CP^2 
⇒ AB^2 = AC^2 + BP^2 - CP^2 
⇒ AB^2 = AC^2 + BP^2 - (AC-AP)^2 
⇒ AB^2 = AC^2+ BP^2 - AC^2 - AP^2 + 2.AC.AP
⇒ AB^2 = BP^2 - AP^2 + 2AC.AP         --------(3)

we also know ,
AP.CP =BP.DP-----(4)

Putting in (3)
AB^2 = BP^2 - AP^2 + AC.AP + AC.AP
          =BP^2-AP(AP-AC)+AC.AP
          =BP^2 -AP.CP +AC.AP
 
AB^2 = BP^2 -BP.DP+AC.AP

AB^2=BP (BP-DP) +AC.AP

AB^2=BP.BD +AC.AP