Question

In triangleABC angleC is a right angle. A semicircle is drawn with AB
as diameter. D is any point of AC produced.
Prove that AB2 = AC AD + BE BD [Fig. 15.105).​

Answer 1

Answer:

Step-by-step explanation:

In Triangle ABC, <ACB = 900 Þ <BCP = 900

So, In Triangle BPC:  BC2 = BP2 –  CP2

Now, In Triangle ABC,

AB2   =  AC2 +  BC2 Þ 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