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Given: ABCD is a square.
Prove: AC ⊥ BD.
Square A B C D is shown. Diagonals are drawn from point A to point C and from point B to point D and intersect at point E.
We are given that ABCD is a square. If we consider triangle AEB and triangle AED, we see that side
is congruent to side AD because sides of a square are congruent. We know that side AE is congruent to side AE by using the
. Finally, we know that side DE is congruent to side
because the diagonals of a square bisect each other. Therefore, triangle AEB is congruent to triangle AED by
congruency. We see that angle AED and angle AEB are a linear pair, and congruent by CPCTC. Thus, the measure of these angles will be 90°, and diagonal AC is perpendicular to diagonal BD by the
.
Sagot :
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