Question

In the given figure,
A) Prove that triangle ABC is similar to triangle ABP is similar to triangle ACP.
B)
$ap^{2} = bp \times pc$
​

Answer 1

Given :-

• ABC is triangle on which AP _|_ BC
• ∆ABC, ∆ABP , ∆APC

To Froof :-

• $\tt{\triangle\:ABC\sim\:\triangle\:ABP\:\sim\triangle\:APC}$

Solution :-

To Froof similarity between this triangles , at first we have to find the common sides or angles between them. By finding common sides we can proof similarity between this triangles.

Calculation for similarity between ∆ :-

➞ ∆ABC, ∆ABP , and ∆APC

$\tt\small{\mapsto \angle\:ABC=\angle\:ACP=\angle\:ABP\:(opposite\: angles\:are\: equal)}$

$\tt\small{\mapsto AB=AB=AC\:(In\: isoceles\:∆\: sides\: equal)}$

$\tt\small{\mapsto AC=AB=AB\:(In\: isoceles\:∆\: sides\: equal)}$

$\tt\small{\mapsto \therefore\:\triangle\:ABC\sim\:\triangle\:ABP\:\sim\:\triangle\:APC\:(by\:CPCTC)}$

Let's proof now :-

• $\tt\small{AP^2=BP.PC}$

By using Pythagoras theorem :-

• In right angle triangle APC

➞ AP² + PC² = AC²

➞ AP² = AC² - PC² -------(i)

• In right angle triangle APB

➞ AP² + BP² = AB²

➞ AP² = AB² - BP²-------(ii)

• Putting together equation (i) and (ii)

➞ ∆ABP similar ∆ APC

➞ AB/BP = AC/PC

➞ AB × PC = BP × AC

➞ AB × PC = BP × AB

➞ PC = BP

• By replacing something here we get :-

➞ AP² + AP² = AC² - PC² + AB² - BP²

➞ 2AP² = 2AB² - PC² - BP² [AC = AB]

➞ 2AP² = 2AB² - 2PC² [ PC = BP]

➞ AP² = AB² - PC²

➞ PC² = AB² - AP². [AB² = AP² + BP²]

➞ PC² = AP² + BP² - AP²

➞ PC² = BP²

➞ PC = BP ( PC.BP = PC² or BP²)

Hence, AP² = PC.BP Proved :-

Answer 2

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

H² = P² + B²

AC² = AP² + PC²

AP² = AC² - PC²

ΔAPB

H² = P² + B²

AB² = AP² + BP²

AP² = AB² - BP²

Now, By comparing

AC/PC = AB/BP

AC × BP = PC × AB

Cancel AB from both sides

BP = PC

AP² + AP² = AC² - PC² + AB² - BP²

2AP² = AC² - PC² + AB² - BP²

As AC = AB

2AP² = AB² - PC² + AB² - BP²

2AP² = 2AB² - PC² - BP²

Also PC = BP

2AP² = 2AB² - PC² - PC²

2AP² = 2AB² - 2PC²

Divide all sides by 2

2AP²/2 = 2AB² - 2PC²/2

AP² = AB² - PC²

AP² - AB² = PC²

Now

• AB² = AP² + BP²

AP² - AP² + BP² = PC²

BP² = PC²

Root both side

BP = PC