MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Solution of Triangle

Let ABC be a ...

Question

Let ABC be a triangle with incentre I and r. Let D, E, F be the feet of the perpendiculars from I to the sides BC, CA and AB respectively, If $r_{1}, r_{2}, a n d r_{3}$ are the radii of circles inscribed in the quadrilaterals AFIE, BDIF and CEID respectively, prove that
$\frac{r_{1}}{r - r_{1}} + \frac{r_{2}}{r - r_{2}} + \frac{r_{3}}{r - r_{3}} = \frac{r_{1} r_{2} r_{3}}{(r - r_{1}) (r - r_{2}) (r - r_{3})}$

A

True

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

B

False

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A True
Let $M N = r_{3} = M P = M Q, I D = r$
$\Rightarrow I P = r - r_{3}$
Clearly IP and IQ are tangents to circle with centre M.
$∴$ IM must be the $∠$ bisector of $∠ P I Q$
$∴$ $∠ P I M = ∠ Q I M = θ_{1}$
Also from $Δ I P M, t a n θ_{1} = \frac{r_{3}}{r - r_{3}} = \frac{M P}{I P}$

Here DI=r
Similarly, in other quadrilaterals, we get
$t a n θ_{2} = \frac{r_{2}}{r - r_{2}}$ and $t a n θ_{3} = \frac{r_{1}}{r - r_{1}}$
Also $2 θ_{1} + 2 θ_{2} + 2 θ_{3} = 2 π \Rightarrow θ_{1} + θ_{2} + θ_{3} = π$
$\Rightarrow t a n θ_{1} + t a n θ_{2} + t a n θ_{3} = t a n θ_{1} . t a n θ_{2} . t a n θ_{3}$
$\Rightarrow \frac{r_{1}}{r - r_{1}} + \frac{r_{2}}{r - r_{2}} + \frac{r_{3}}{r - r_{3}} = \frac{r_{1} r_{2} r_{3}}{(r - r_{1}) (r - r_{2}) (r - r_{3})}$

Suggest Corrections

thumbs-up

0

Similar questions

A B C

be a triangle with in centre

I

r

D, E, F

be the feet of the perpendiculars from

I

B C, C A

A B

respectively. If

r_{1}, r_{2} r_{3}

are the radii of circles inscribed in then prove

\frac{r_{1}}{r - r_{1}}_\frac{r_{2}}{r - r_{2}} + \frac{r_{3}}{r - r_{3}} = \frac{r_{1} r_{2} r_{3}}{(r - r_{1}) (r - r_{2}) (r - r_{3})}

Q. Assertion :In a

△ A B C

a < b < c

r

is inradius and

r_{1}, r_{2}, r_{3}

are the axradii opposite to angle

A, B, C

respectively, then

r < r_{1} < r_{2} < r_{3}

△ A B C, r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = \frac{r_{1} r_{2} r_{3}}{r}

r_{1}, r_{2}, r_{3}

are the radii of the escribed circles of a triangle

A B C

r

is the radius of its incircle,then

r_{1} r_{2} r_{3} - r (r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1})

Δ

A B C

the sides opposite to angles

A, B, C

a, b, c

respectively. Let

r_{1} . r_{2}

r_{3}

be the ex-radii of the traingle and

r

be the in-radius.

Find the value of

r_{1} + r_{2} + r_{3} - r

Q. Prove that :

a (r r_{1} + r_{2} r_{3}) = b (r r_{2} + r_{3} r_{1}) = c (r r_{3} + r_{1} r_{2})

r

is inradius and

r_{1}, r_{2}, r_{3}

are exradius of triangle

A B C

a, b, c

are the corresponding sides.

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Solution of Triangle

Standard XI Mathematics

Join BYJU'S Learning Program

CrossIcon