Grade 1010 grade maths

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= ACAP+ BDBP

Palak Tiwari

8 Years agoGrade 10

2 Answers

Arun

Approved Tutor Answer8 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

8 Years ago

How are they similar the triangles?

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