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3 years ago

In the figure below, AB=AE, AC=AD and AP is perpendicular to BE. Prove that BC=DE.

Mathematics

1 answer:

NISA [10]3 years ago

5 0

Step-by-step explanation:

1) In the figure, as AB is equal to AE, ABE is an equilateral triangle.

As AP is perpendicular to BE

=> AP is the height of the triangle ABE.

In an equilateral triangle, the median and the height is the same, so that AP is also the median of the triangle.

=> P is the midpoint of BE

=> PE = PB

2) In the figure, as AC = AD, so that ACD is an equilateral triangle.

As AP is perpendicular to BE, so that it is perpendicular to CD as well

=> AP is the height of the triangle ACD

In an equilateral triangle, the median and the height is the same, so that AP is also the median of the triangle ACD.

=> P is the midpoint of CD

=> PC = PD

We have:

+) PB = PE

+) PC = PD

=> PB - PC = PE - PD => BC = DE

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In all the equations we have solved so far, all the variable terms were on only one side of the equation with the constants on the other side. This does not happen all the time—so now we will learn to solve equations in which the variable terms, or constant terms, or both are on both sides of the equation.

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