Question

(1) A line cuts two sides AB and AC of ABC in points P and
then,
(a) Draw seg BQ and write ALAPO)
AC LABO)
(b) Write ALL ABO)
ACAABC)
(c) Multiply the result obtained in (a) and (b)
and prove LAAPQ) AP X AQ
АС ДАВС) AB x AC
nº consider BAC​

Answer 1

SOLUTION :  

Given : In ∆ABC, in which ∠A is obtuse, PB ⊥ AC and QC ⊥ AB.

In ∆APB and ∆AQC  

∠P = ∠Q   (each 90°)

∠PAB = ∠QAC   (vertically opposite angle)

∆APB ~ ∆AQC  

[By AA similarity]

∴ AP/AQ = AB/AC

[Corresponding parts of similar ∆ are proportional]

AP × AC = AQ × AB ………..(1)

(ii) In ∆BPA ,

AB² = BP² + PA²

[By using Pythagoras theorem]

BP² = AB² - AP² ………….(2)

In ∆BPC, by pythagoras theorem

BC² = BP² + PC²

[By using Pythagoras theorem]

BC² = AB² - AP² +(AP+ AC)²

[From eq 2]

BC² = AB²  - AP² + AP² + AC² +2AP×AC

[(a+b)² = a² + b² +2ab]

BC² = AB²  + AC² +2AP×AC …………(3)

In ∆AQC,  

AC² = AQ² + QC²

[By using Pythagoras theorem]

QC² = AC² - AQ² …………..(4)

In ∆BQC,  

BC² = CQ² + BQ²

[By using Pythagoras theorem]

BC² = AC² - AQ² + (AB + AQ)²

[From eq 4]

BC² = AC² - AQ² + AB² + AQ² + 2AB×AQ

[(a+b)² = a² + b² +2ab]

BC² = AC² + AB²  + 2AB×AQ …………(5)

On Adding eq 3 & 5

BC² = AB²  + AC² +2AP×AC

BC² = AC² + AB²  + 2AB×AQ

-----------------------------------------

2BC² = 2AC² +2AB² +2AP×AC + 2AB×AQ

2BC² = 2AC² +2AP×AC + +2AB² +2AB×AQ

2BC² = 2AC[AC+  AP] + 2AB[AB + AQ]

2BC² = 2AC × PC  + 2AB × BQ

[PC = AC +AP & BQ = AB + AQ]

2BC² = 2(AC × PC  + AB × BQ)

BC² = (AC × PC  + AB × BQ)    [Divide by 2 ]  

HOPE THIS ANSWER WILL HELP YOU…