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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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