Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Prove that AC + AD = BC.

ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Prove that   AC + AD = BC.

Answers (1)

Given, ABC is right angle triangle

AB = AC, CD is bisector at AB

To prove: AC + AD = BC

Proof: \becauseAB = AC (Given)

By Pythagoras theorem 

BC2 =AB2 + AC2 = AC2 + AC2 = AC2 + AC2              (\because AB = AC)

BC =\sqrt{2}AC                                                                                           

By angle bisector theorem:

\frac{AD}{BD}=\frac{AC}{BC}=\frac{AC}{\sqrt{2}AC}=\frac{1}{\sqrt{2}}\left ( From \;above \right )

Let AB = a, AD = b

\frac{b}{a-b}=\frac{1}{\sqrt{2}}\\ \Rightarrow b=\frac{a}{1+\sqrt{2}}\\ \Rightarrow b=a\left ( \sqrt{2}-1 \right )\\ \Rightarrow a+b=a\sqrt{2}

AC + AD = BC                        (\because AB = AC)

 

\Rightarrow \frac{b}{a-b}-\frac{a}{a\sqrt{2}}=a\left ( \sqrt{2}-1 \right )\\ \Rightarrow a+b=a\sqrt{2}

 

We know, AC = a, AD = b, BC = a\sqrt{2}

=> AC + AD = BC

Hence proved

Posted by

infoexpert24

View full answer

Similar Questions

Latest Question