Geometry - congruent triangles

The triangle ABC has the measure of 90 degrees of the angle A, the bisector of the angle ABC intersects in point D the side AC and DE is the altitude of the triangle BDC. Find that:

a)  AE is perpendicular to BD ; 

b)  AE is parallel to FC,  where F is the intersecting point of AB and DE: 
 {F} = AB ∩ DE.

Solution:

Analyzing the problem, we find that the hypothesis provides:

 (1) the triangle ABC has the measure of 90 degrees of the angle A; 

(2) BD the bisector of the angle ABC ;

(3) DE is the altitude of the triangle BDC.

From these data we find the following:

From the relation (1) : the triangle ABC is a right triangle with the right angle BAC.

We draw triangle ABC and mark on the figure the right angle with a sign „ ┐”.

From the relation (2), the angle bisector BD divides the angle ABC into two equal (congruent) parts:

the measure of the angle ABD = the measure of the angle DBC

m (∡ABD) = m (∡ DBC)  (4)

we mark on the figure small arcs for these congruent angles.

From the relation (3) results the DE is perpendicular to BC.

We mark on the figure this perpendicularity „ ┐”.

 

 

We look the figure and notice the triangles ABD and EBD.

These triangle have some common elements.

The triangles ABD and EBD are right triangles, the side BD is a common side, and the angles ABD and DBE are congruent => the angles ADB and EDB are also congruent,

m(∡ADB) = 90 – m(∡ABD) = 90- m(∡DBE) = m(∡EDB).

The triangles ABD and EBD are congruent (case Angle – Side – Angle).

From this congruence results all elements of these triangles are also congruent. We write the sides congruent:

AD = DE   (5)

BA = BE   (6)

From the relation (6)  results : the triangle ABE is an isosceles triangle.

The unequal side of the isosceles triangle ABE is AE referred to as the 'base' of the triangle.

The base angles of an isosceles triangle are always equal. In the figure above, the angles ∡BAE and ∡BEA are always the same.

We note the point M of intersection of side AE and BD.

∡BAE = ∡BEA, or ∡BAM = ∡BEM  (7)

We see the triangles BAM and BEM:

BM – a common side,

∡BAM = ∡BEM  and

from relation (4) m (∡ABD) = m (∡ DBC)  we can write using the point M:
m (∡ABM) = m (∡ MBC).

Conclusion: the triangles BAM and BEM are congruent by the case Angle – Side -Angle.

From this congruence results the congruent angles BMA and BME.

But adding the measures of these angles we obtain a straight angle with 180 degrees measure.

m(∡BMA) = m(∡BME) = 180/2 = 90 degrees.

ð BM ⊥ AE , BM is altitude of the triangle ABE.

Because the points B, M and D are collinear we can write the last relation:

BD ⊥ AE.  "what was to be demonstrated (q.e.d.)",  

b) In the triangles BFC anb ABC we have

the side CA Pis perpendicular to FB ,
the side FE is perpendicula to BC,  CA ∩ FE = {D} (the point D).

The right  triangles BEF and ABC have also BA = BE   (6) and the common angle B, results

m(∡EFB)=m(∡ACB) = 90 degrees - m(∡B)

and the right  triangles BEF and ABC are congruent by the case  Angle –Side – Angle.

ð BF = BC, 

From this congruence,  the triangle BFC is an isosceles triangle with the unequal angle B,

ð m(∡BFC) = m(∡BCF) = (180⁰ – m(∡B))/2,

The triangle ABE is also a isosceles triangle with the unequal angle B, ∡BAE = ∡BEA (relation 7)

ð m(∡BAE) = m(∡BEA) = (180⁰ – m(∡B))/2,

ð  m(∡BFC) = m(∡BCF) = m(∡BAE) = m(∡BEA) = (180⁰ – m(∡B))/2,

ð AE is parallel to FC (m(∡BFC) =  m(∡BAE)= (180⁰ – m(∡B))/2 = Converse of the corresponding angles postulate: if any two lines are cut be a transversal and the corresponding angles are congruent, then the two lines are parallel).

What was to be demonstrated (q.e.d.).