*Notes to a video lecture on http://www.unizor.com*

__Geometry+ 06__

*Problem A*

Given an isosceles triangle Δ

*with*

**ABC***and ∠*

**AB=BC***°.*

**ABC=20**Point

*on side*

**D***is chosen such that ∠*

**BC***°.*

**CAD=60**Point

*on side*

**E***is chosen such that ∠*

**AB***°.*

**ACE=50**Find angle ∠

*.*

**ADE***Solution A*

*Problem B*

Given two triangles Δ

*and Δ*

**A**_{1}B_{1}C_{1}*with the following properties:*

**A**_{2}B_{2}C_{2}(a) side

*of the first triangle equals to side*

**A**_{1}B_{1}*of the second;*

**A**_{2}B_{2}(b) angles opposite to these sides, ∠

*and ∠*

**A**_{1}C_{1}B_{1}*, are equal to each other;*

**A**_{2}C_{2}B_{2}(c) bisectors of these angles,

*and*

**C**_{1}X_{1}*, are also equal to each other.*

**C**_{2}X_{2}Prove that these triangles are congruent.

*Proof*

Put side

*on top of*

**A**_{1}B_{1}*and draw a circle around Δ*

**A**_{2}B_{2}*.*

**A**_{1}B_{1}C_{1}Obviously, point

*would lie on this circle because of equality of angles opposite to equal sides.*

**C**_{2}Assume that points

*and*

**C**_{1}*do not coincide.*

**C**_{2}Extend bisectors

*and*

**C**_{1}X_{1}*to intersection with a circle at point*

**C**_{2}X_{2}*. It must be the same point for both bisectors because each divides the arc ⌒*

**P***in half.*

**A**_{1}B_{1}First, let's prove that triangle Δ

*is similar to triangle Δ*

**PC**_{1}C_{2}

**PX**_{1}X_{2}Our proof will be based on congruency of two angles of one triangle to two corresponding angles of another.

For brevity, when we use an expression "angle is measured by an arc (or a fraction of it)", we mean that this angle is equal to a central angle subtended (supported) by this arc (or a fraction of it).

We will also use known theorems of Geometry:

(

*Theorem 1*) An angle inscribed into a circle (that is, formed by two chords with one common point on a circle) is measured by half an arc it cuts from a circle by its two rays.

(

*Theorem 2*) An angle formed by two chords intersecting inside a circle is measured by half of a sum of two opposite arcs these chords cut.

These theorems were presented in

*UNIZOR.COM - Math 4 Teens - Geometry - Circles - Mini Theorems 1*and

*...Mini Theorems 2*.

1. ∠

*is measured by half of sum of arcs ⌒*

**PC**_{1}C_{2}*and ⌒*

**C**_{2}B_{2}*;*

**B**_{2}P∠

*is measured by half of sum of arcs ⌒*

**PX**_{2}A_{2}*and ⌒*

**C**_{2}B_{2}*;*

**PA**_{2}but arcs ⌒

*and ⌒*

**B**_{2}P*are equal, from which follows that*

**PA**_{2}∠

*equal to ∠*

**PC**_{1}C_{2}*.*

**PX**_{2}A_{2}2. ∠

*is measured by half of sum of arcs ⌒*

**PC**_{2}C_{1}*and ⌒*

**C**_{1}A_{2}*;*

**A**_{1}P∠

*is measured by half of sum of arcs ⌒*

**PX**_{1}B_{1}*and ⌒*

**C**_{2}B_{2}*;*

**PA**_{2}but arcs ⌒

*and ⌒*

**B**_{2}P*are equal, from which follows that*

**PA**_{2}∠

*equal to ∠*

**PC**_{1}C_{2}*.*

**PX**_{2}A_{2}Since two angles of Δ

*are congruent to two angles of Δ*

**PC**_{1}C_{2}*, these triangles are similar.*

**PX**_{1}X_{2}Consequently, their sides are proportional

**PC**_{1}/PX_{2}= PC_{2}/PX_{1}If angle bisectors

*and*

**C**_{1}X_{1}*are equal, segments*

**C**_{2}X_{2}*and*

**C**_{1}P*must be equal to preserve the proportionality of corresponding sides of triangles Δ*

**C**_{2}P*and Δ*

**PC**_{1}C_{2}*.*

**PX**_{1}X_{2}Indeed, if

*,*

**C**_{1}X_{1}=C_{2}X_{2}=d*and*

**PX**_{1}=x_{1}*, then*

**PX**_{2}=x_{2}

**(d+x**_{1})/x_{2}= (d+x_{2})/x_{1}**⇒**

**d·x**_{1}+x_{1}² = d·x_{2}+x_{2}²**⇒**

**d·(x**_{1}−x_{2}) = (x_{2}−x_{1})·(x_{1}+x_{2})**⇒**

*, because otherwise*

**x**_{1}= x_{2}*would be negative*

**d***.*

**−(x**_{1}+x_{2})From

*follows*

**C**_{1}P = C_{2}P**⇒**⌒

*⌒*

**C**_{1}A_{1}P =

**C**_{2}B_{2}P**⇒**⌒

*⌒*

**C**_{1}A_{1}=

**C**_{2}B_{2}**⇒**

**C**_{1}A_{1}= C_{2}B_{2}Analogously,

**⇒**

**C**_{1}B_{1}= C_{2}A_{2}which proves that triangles Δ

*and Δ*

**A**_{1}B_{1}C_{1}*are congruent by three sides.*

**A**_{2}B_{2}C_{2}
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