Question

In the given figure , triangle ABC and triangle CPD .Prove :- BP ×PD = EP × PC​

Answer 1

Given :-

• A triangle with sides ABC

To Prove :-

• A) Prove that triangle ABC is similar to triangle ABP is similar to triangle ACP.

• B) AP² = BP × PC

Solution :-

• 1]AB = AC

Also

∠ABC = ∠ABP = ∠ACP

2]Since we may see it is a right angled triangle.

By using pythagoras theorem in

ΔAPB

$\longrightarrow$H² = P² + B²

$\longrightarrow$AC² = AP² + PC²

$\longrightarrow$AP² = AC² - PC²

ΔAPB

$\longrightarrow$H² = P² + B²

$\longrightarrow$AB² = AP² + BP²

$\longrightarrow$AP² = AB² - BP²

Now, By comparing

$\longrightarrow$AC/PC = AB/BP

$\longrightarrow$AC × BP = PC × AB

$\longrightarrow$Cancel AB from both sides

$\longrightarrow$BP = PC

$\longrightarrow$AP² + AP² = AC² - PC² + AB² - BP²

$\longrightarrow$2AP² = AC² - PC² + AB² - BP²

$\longrightarrow$As AC = AB

$\longrightarrow$2AP² = AB² - PC² + AB² - BP²

$\longrightarrow$2AP² = 2AB² - PC² - BP²

$\longrightarrow$Also PC = BP

$\longrightarrow$2AP² = 2AB² - PC² - PC²

$\longrightarrow$2AP² = 2AB² - 2PC²

Divide all sides by 2

$\longrightarrow$2AP²/2 = 2AB² - 2PC²/2

$\longrightarrow$AP² = AB² - PC²

$\longrightarrow$AP² - AB² = PC²

Now

$\longrightarrow$AB² = AP² + BP²

$\longrightarrow$AP² - AP² + BP² = PC²

$\longrightarrow$BP² = PC²

Root both side

$\longrightarrow$BP = PC

Answer 2

mei Mohammad zakariya hu

hlo kuch yaad aaya

insta pe h

to

meri id zakariya_khan10