Basic and Rich Geometry Concepts Part 8, Internal Angle bisectors and Segment Ratios at Incentre of a triangle | SureSolv

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Basic and Rich Geometry Concepts Part 8, Internal Angle bisectors and Segment Ratios at Incentre of a triangle

Incentre, Internal angle bisectors and Ratio Relations of Segmentation by angle bisectors as well as at incentre of a triangle

geometry-concepts-8-segmentation-ratio-at-incentre

Incentre of a triangle is the centre of the inscribed circle, and also the meeting point of the three internal bisectors.

In this 8th concept tutorial session on Geometry, these aspects will dealt with first with proofs, and then we'll go through the relations between the sides of the triangle and the segmentation at the incentre as well as the segments made by the angle bisectors on the opposite sides.

Incentre, the centre of the inscribed circle of a triangle, and the internal angle bisectors

Incentre of a triangle is the centre of the circle inscribed in it. The centre O of the circle inscribed in the $\triangle ABC$ in figure below is the incentre of the triangle.

basic-rich-geometry-concepts-8-incentre-angle-bisectors-segments-1

OP, OQ and OR are the radii of the incircle joining its centre and the points at which they are perpendicularly dropped to the tangents AB, BC and CA. P, Q and R are the tangent points and AB, BC and CA are the three sides of the $\triangle ABC$.

This describes the main elements we will use now.

The incircle with centre O is defined and drawn then as,

The circle drawn inside a triangle to which each of the three sides are tangents to the inscribed circle. The centre of this circle is the incentre of the triangle.

Incentre and Internal angle bisectors of a triangle

The bisectors of the internal angles, $\angle A$, $\angle B$ and $\angle C$, the line segments AO, BO and CO of the triangle, are the internal angle bisectors of $\triangle ABC$. Basic property of these internal angle bisectors is,

The three internal angle bisectors of a triangle meet at a single point which is the incentre of the triangle.

Proof of internal angle bisectors meeting at incentre

The same graphic as above we will use for the proof.

basic-rich-geometry-concepts-8-incentre-angle-bisectors-segments-1

In the two triangles $\triangle AOP$ and $\triangle AOR$, radii OP and OR dropped on sides AB and AC to the tangent points P and Q respectively are perpendicular to the tangents AB and AC.

So, $\angle APO=\angle ARO=90^0$.

The two triangles are then right-angled and hypotenuse AO common with $OR=OP$ as these are radii. This makes the third pair of sides also equal, $AP=AR$.

With three pairs of sides equal, the two triangles, $\triangle AOP$ and $\triangle AOR$ are congruent.

It follows, all the corresponding angles of the two triangles are equal. So,

$\angle PAO=\angle RAO$.

That is, AO is the internal angle bisector of $\angle A$ of $\triangle ABC$.

In exactly the same way, the other two line segments, BO and CO are the internal angle bisectors of the other two angles $\angle B$ and $\angle C$.

With this result. the point O can be observed in two ways as,

1. Centre of the circle inscribed in $\triangle ABC$, and,

2. Concurrent intersection point of three internal angle bisectors of $\triangle ABC$.

An internal angle bisector segments the opposite side in the ratio of the two adjacent sides and its proof

The first segmentation ratio relation in a triangle with incentre states,

An internal angle bisector of a triangle segments the side opposite to the angle bisected in the ratio of the other two sides adjacent to the angle bisected.

For explaining the proof of the result, we will use the following graphic.

basic-rich-geometry-concepts-8-incentre-angle-bisectors-segments-2

By this segmentation result in $\triangle ABC$,

$\displaystyle\frac{BE}{CE}=\frac{AB}{AC}$.

Let us see how this is true.

Draw a line parallel to the internal angle bisector AE passing through B. It meets side CA extended at S.

As in $\triangle BCS$, SB is parallel to AE, the two triangles $\triangle BCS$ and $\triangle ECA$ are similar.

Reason: $\angle C$ common and the other two pairs of angles also equal as $SB || AE$ so that A-A-A similarity condition is satisfied.

As a result, ratio of corresponding sides are equal,

$\displaystyle\frac{SC}{AC}=\frac{BC}{CE}$.

Subtract 1 from both sides with the result,

$\displaystyle\frac{SC-AC}{AC}=\frac{BC-CE}{CE}$,

Or, $\displaystyle\frac{SA}{AC}=\frac{BE}{CE}$.

But as $SB || AE$, 

$\angle SBA=\angle BAE$, and,

$\angle ASB=\angle CAE$.

So with two parts of bisected $\angle A$, $\angle BAE=\angle CAE$,

$\angle SBA =\angle ASB$.

As a result, $\triangle ASB$ is isosceles, and, $SA = AB$.

The segment ratio takes its final form then as,

$\displaystyle\frac{BE}{CE}=\frac{AB}{AC}=\frac{c}{b}$, using the short form of naming the sides of a triangle.

In the same way,

$\displaystyle\frac{CF}{AF}=\frac{BC}{AB}=\frac{a}{c}$, and,

$\displaystyle\frac{AD}{BD}=\frac{AC}{BC}=\frac{b}{a}$.

This is a relation between ratio of two sides of the triangle and the ratio of segments of the third side made by an internal angle bisector.

Ratio of segments of an internal angle bisector at the incentre in terms of sides of the triangle

This segmentation ratio in a triangle with incentre relates the ratio of the segments of an internal bisector at the incentre with the sides of the triangle.

Specifically this relation states,

The ratio of the segments of an internal bisector made at the incentre is equal to the ratio of the sum of the adjacent sides to the opposite side.

With respect to $\triangle ABC$ in the following diagram, this relation is expressed as, 

$\displaystyle\frac{CO}{OD}=\frac{AC+BC}{AB}=\frac{a+b}{c}$.

basic-rich-geometry-concepts-8-incentre-angle-bisectors-segments-3

Let us go through the proof of this result.

Draw a line segment DT parallel to AO that meets the extended side CA at T.

By the construction the two triangles, $\triangle CTD$ and $\triangle CAO$ are similar so that the ratio of the corresponding sides of the two triangles are equal. For instance,

$\displaystyle\frac{CD}{CO}=\frac{CT}{AC}$.

Subtract 1 from both sides with the result,

$\displaystyle\frac{CD-CO}{CO}=\frac{CT-AC}{AC}$,

Or, $\displaystyle\frac{OD}{CO}=\frac{AT}{AC}$.

Now, as $\angle ADT=\angle DAO$ and $\angle ATD= \angle CAO$ by the parallelism of DT and AO, as well as, $\angle DAO= \angle CAO$ as these are two equal bisected parts of $\angle CAD$, we have,

$\angle ADT=\angle ATD$ and as a result $\triangle ADT$ isosceles.

It follows, sides, $AT=AD$.

Substitute this result into the segment relation to get the result,

$\displaystyle\frac{OD}{CO}=\frac{AD}{AC}$,

Or, $\displaystyle\frac{CO}{OD}=\frac{b}{AD}$.

IIn the RHS, $AD$ being a segment of the side $AB$, objective now is to eliminate its reference in terms of the other two sides. We have to form a relation equating $\displaystyle\frac{b}{AD}$ with an expression in terms of only the sides of the triangle.

Recall our earlier result of segmentation at the opposite side result,

$\displaystyle\frac{AD}{BD}=\frac{b}{a}$.

Invert this to get,

$\displaystyle\frac{BD}{AD}=\frac{a}{b}$.

Add 1 to both sides of the equation,

$\displaystyle\frac{AD+BD}{AD}=\frac{a+b}{b}$,

Or, $\displaystyle\frac{AB}{AD}=\frac{c}{AD}=\frac{a+b}{b}$,

Or, $\displaystyle\frac{b}{AD}=\frac{a+b}{c}$.

Substitute this desired result into the segmentation result to eliminate side segment AD. The final desired form of segmentation ratio equality at incentre is then,

$\displaystyle\frac{CO}{OD}=\frac{b}{AD}=\frac{a+b}{c}$.

The ratio of the segments of internal angle bisector at incentre equals the ratio of the sum of adjacent sides and the opposite side.

In the same way, we get the other two rich segmentation results for internal angle bisectors at incentre as,

$\displaystyle\frac{AO}{OE}=\frac{b+c}{a}$, and,

$\displaystyle\frac{BO}{OF}=\frac{c+a}{b}$.

This is the final form of rich segmentation ratio result for internal angle bisectors that is often used for solving difficult Geometry problems quickly in a few simple steps.

We will solve such an example problem now to show how this result makes possible solution of an otherwise difficult problem easy.

Problem example on segmentation of internal angle bisectors at the incentre

Question: In the given figure, O is the incentre of $\triangle ABC$. If $\displaystyle\frac{AO}{OE}=\frac{5}{4}$ and $\displaystyle\frac{CO}{OD}=\frac{3}{2}$, what is the value of $\displaystyle\frac{BO}{OF}$?

basic-rich-geometry-concepts-8-incentre-angle-bisectors-segments-4

  1. $\displaystyle\frac{19}{14}$
  2. $\displaystyle\frac{19}{7}$
  3. $\displaystyle\frac{38}{7}$
  4. $\displaystyle\frac{38}{17}$

Solution to example problem

Given,

$\displaystyle\frac{AO}{OE}=\frac{b+c}{a}=\frac{5}{4}$,

Or, $\displaystyle\frac{a+b+c}{a}=\frac{9}{4}$, adding 1 to both sides.

Inverting,

$\displaystyle\frac{a}{a+b+c}=\frac{4}{9}$.

The other given relation similarly is,

$\displaystyle\frac{CO}{OD}=\frac{a+b}{c}=\frac{3}{2}$,

Or, $\displaystyle\frac{a+b+c}{c}=\frac{5}{2}$, adding 1 to both sides of the equation.

Inverting in the same way as before,

$\displaystyle\frac{c}{a+b+c}=\frac{2}{5}$.

Add these two intermediate results,

$\displaystyle\frac{c+a}{a+b+c}=\displaystyle\frac{2}{5}+\displaystyle\frac{4}{9}=\displaystyle\frac{38}{45}$.

Invert this result again,

$\displaystyle\frac{a+b+c}{c+a}=\frac{45}{38}$.

Subtract 1 from both sides,

$\displaystyle\frac{b}{c+a}=\frac{7}{38}$.

Inverting we get the solution.

$\displaystyle\frac{c+a}{b}=\frac{BO}{OF}=\frac{38}{7}$.

Answer: Option c: $\displaystyle\frac{38}{7}$.

This is use of rich concept of segmentation ratio of internal angle bisectors at incentre and use of smart algebraic techniques.


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