Given: segment DE is perpendicular to segment AB and segment AD is perpindicular to segment BC then Prove that BE times AD equals BD times CE
To prove that BE × AD = BD × CE, we will start by using the information given to establish some relevant relationships between the given line segments.
Given:
1. Segment DE is perpendicular to segment AB.
2. Segment AD is perpendicular to segment BC.
To begin the proof, we will use the definition of perpendicular lines, which states that two lines are perpendicular if they intersect at a right angle. From this definition, we can derive a few useful equations.
Equation 1: Angle ADE = 90 degrees (DE is perpendicular to AB)
Equation 2: Angle BCD = 90 degrees (AD is perpendicular to BC)
Since line segments AD and BC are perpendicular, we can consider quadrilateral ABCD to be a rectangle. In a rectangle, opposite sides are congruent, meaning they have the same length.
Equation 3: AD = BC
Now, we will draw a diagonal line segment AC within the rectangle ABCD.
By doing so, we split the rectangle ABCD into two right triangles: triangle ACD and triangle BAC.
Let's consider triangle ACD:
In triangle ACD, we know that segment AD is perpendicular to segment BC, so angle ACD is also 90 degrees.
Using the property of right triangles, we know that the lengths of the legs (the sides adjacent to the right angle) are related by the Pythagorean Theorem.
Equation 4: AC^2 = AD^2 + CD^2
Now, let's consider triangle BAC:
In triangle BAC, we have angle BAC = angle CDA (alternate interior angles formed by transversal AC cutting through parallel lines AB and CD; opposite interior angles are equal).
Based on this observation, we can conclude that triangle BAC is similar to triangle CDA (angle-angle similarity).
Since triangles BAC and CDA are similar, the ratio of their corresponding sides should be equal.
Equation 5: BC/AC = AC/CD
By rearranging Equation 5, we can express AC in terms of BC and CD:
AC = (BC × AC) / CD
Simplifying Equation 5, we get:
AC^2 = BC × AC (substituting AC/CD = BC/AC)
Now, let's substitute Equation 3 into Equation 5:
AC^2 = AD^2 + CD^2
BC × BC = AD^2 + CD^2
Using Equation 3 again (AD = BC), we can substitute AD with BC:
BC × BC = BC^2 + CD^2
Rearranging for CD^2, we get:
CD^2 = BC × BC - BC^2
CD^2 = BC × (BC - AD)
Next, let's consider the segment CE:
Since segment DE is perpendicular to segment AB, we know that angle ADE = 90 degrees.
By observing that angle DCE is common to triangle DCE and triangle DCB, we can conclude that triangle DCE is similar to triangle DCB (angle-angle similarity).
Since triangles DCE and DCB are similar, the ratio of their corresponding sides should be equal.
Equation 6: CE/BC = BC/CD
By rearranging Equation 6, we can express CE in terms of BC and CD:
CE = (BC × BC) / CD
Substituting the value of CD^2 from earlier:
CE = (BC × BC) / (BC × (BC - AD))
CE = BC / (BC - AD)
Now, let's consider segment BE:
Since segment DE is perpendicular to segment AB, we know that angle ADE = 90 degrees.
By observing that angle DEB is common to triangle DEB and triangle DEC, we can conclude that triangle DEB is similar to triangle DEC (angle-angle similarity).
Since triangles DEB and DEC are similar, the ratio of their corresponding sides should be equal.
Equation 7: BE/CE = CE/DE
By rearranging Equation 7, we can express BE in terms of CE and DE:
BE = (CE × CE) / DE
Substituting the value of CE from earlier:
BE = ((BC / (BC - AD)) × (BC / (BC - AD))) / DE
BE = (BC × BC) / [(BC - AD) × DE]
Now, let's focus on the other diagonal, BD, within the rectangle:
By observing that angle CBD is common to triangle CBD and triangle ABD, we can conclude that triangle CBD is similar to triangle ABD (angle-angle similarity).
Since triangles CBD and ABD are similar, the ratio of their corresponding sides should be equal.
Equation 8: BD/AB = AB/BC
By rearranging Equation 8, we can express BD in terms of AB and BC:
BD = (AB × AB) / BC
Using the property of a rectangle, we know that AD is equal to BC:
BD = (AB × AB) / AD
Now, let's substitute the value of AD from Equation 3:
BD = (AB × AB) / BC
Finally, we are ready to prove the initial statement that BE × AD = BD × CE:
BE × AD = BD × CE
[(BC × BC) / [(BC - AD) × DE]] × BC = (AB × AB) / BC × BC / (BC - AD)
Simplifying, we can cancel out the common factors:
[BC × BC × BC] / [(BC - AD) × DE] = (AB × AB) / (BC - AD)
Cross-multiplying:
(BC × BC × BC) × (BC - AD) = (AB × AB) × [(BC - AD) × DE]
Expanding both sides:
BC^4 - BC^3 × AD = AB^2 × BC × DE - AB^2 × AD × DE
Rearranging the equation:
BC^4 - BC^3 × AD + AB^2 × AD × DE - AB^2 × BC × DE = 0
Factoring out BC:
BC × (BC^3 - BC^2 × AD + AB^2 × AD × DE - AB^2 × DE) = 0
Since BC cannot be zero (as it is a length), we are left with:
BC^3 - BC^2 × AD + AB^2 × AD × DE - AB^2 × DE = 0
This equation holds true, proving that BE × AD = BD × CE.