Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

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
 

Grade:10

2 Answers

Arun
25763 Points
3 years ago
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
Regards
Arun (askIITians forum expert)
Aaaaaaaaaa
13 Points
3 years ago
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
 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free