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Grade: 10
In a triangle ABC, a semicircle is drawn with AB as diameter and circle passing through C. P is a point on AC produced . When joined BP intersects the semicircle at D. Prove that-   AB^2= AC*AP+ BD*BP
one year ago

Answers : (2)

22796 Points
In Triangle ABC,  
So, In Triangle BPC:  BC2 = BP2 –  CP2 
Now, In Triangle ABC,
AB2   =  AC2 +  BC2 and  AB2 = AC2 +  BP2 –  CP2    
AB2 = AC2 – CP2 + BP2 
AB2 = (AC - CP) (AC + CP) + BP2
AB2 = (AC - CP) (AP) + BP2 
AB2 = (AC x AP -  CP x AP) + BP2
AB2 = AC x AP + BP2  -  CP x AP
But,  AP x CP = BP x PD (D APB ~ D CPB)
AB2 = AC x AP + BP2 - BP x PD
AB2 = AC x AP + BP(BP – PD)
AB2 = AC x AP + BP x BD
Hence Proved
Arun (askIITians forum expert)
one year ago
13 Points
How are they similar the triangles?
I would appreciate if you could reply at the earliest because I have an exam tomorrow and you system wouldn’t let me post my query /answer if the comment is less than 100 char
one year ago
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