Question

If the inscribed circle of triangle ABC touches BC at D. Prove that AB-BD=AC-CD

Answer 1

Answer:

REFER IN ATTACHMENT

MARK AS BRAINLIST ANSWER PLEASE

Answer 2

Answer:

A circle is inscribed in triangle ABC and the circle touches the triangle at D.

AB + CD = AC + BD

Hence, AB - BD = AC - CD

Step-by-step explanation:

Given,

A circle is inscribed in triangle ABC and the circle touches the triangle at D, E, and F.

AB = AC    -(i)

By using the theorem (tangents from an external point to a circle are equal)

So,

AF = AE   -(ii)

BF = BD    -(iii)

CD = CE    -(iv)

Now, AB = AC (given)

So, AF + FB = AE + EC

As, AF = AE

so, BF = CE    -(v)

BF = BD & CE = CD (according to the theorem)

So, BD= CD    -(vi)

from equations (i) and (vi), we get

AB + CD = AC + BD

Hence, AB - BD = AC - CD

To learn more about the triangle, click on the link below:

https://brainly.in/question/325089

To learn more about Circle, click on the link below:

https://brainly.in/question/28964

#SPJ3