একটি চতুর্ভুজ সমদ্বিবাহু ট্রাপেজিয়াম প্রমাণ

One way to prove that a quadrilateral is an isosceles trapezoid is to show:

1. The quadrilateral has two parallel sides.

2. The lower base angles are congruent and the upper base angles are congruent.

Let's look at an example to illustrate how one might go about proving that a quadrilateral is an isosceles trapezoid.

First, let's show that $\overline{AD}$ is parallel to $\overline{BC}.$ The line segment that contains $\overline{AB}$ is a transversal that intersects with $\overline{AD}$ and $\overline{BC}.$ Notice that $m\angle BAD$ AND its corresponding angle are both equal to $75°.$ Thus, $\overline{AD} || \overline{BC}.$ For more information on why this is true, here is an explanation on the relationship between transversals, parallel lines, and angles.

Since $\overline{AD} || \overline{BC},$ we know that ABCD is a trapezoid. Next, we have to show that the corresponding base angles are congruent.

If we look at the trapezoid, we can see that $\angle ADC$ is the supplementary angle to $105°.$ To calculate $m\angle ADC,$ we subtract $105°$ from $180°$ to get $75°.$ So: $$m\angle ADC = m\angle BAD = 75°$$ The lower base angles are congruent!

We can also check that the upper base angles are congruent. Because $\overline{AD} || \overline{BC},$ we can conclude that $m\angle BCD = 105°.$ We also know that $\angle ABC$ is supplementary to $75°,$ so: $$m\angle ABC = 180° - 75° = 105°$$ So: $$m\angle BCD = m\angle ABC = 105°$$ The upper base angles are congruent!

We have shown that this quadrilateral has two parallel sides and congruent corresponding base angles. Thus, ABCD is an isosceles trapezoid.

Thus, proving trapezoid $ABCD$ with legs CD, AB is isosceles is equivalent to showing

a) $\angle DAB = \angle CBA$ b) $\angle ADC = \angle BCD$

These all follow from the equal angles condition.

c) $AD = BC$. This follows from the equal legs condition.

d) $\angle ADB = \angle ACB$ or $\angle CAB = \angle CDB$ or similar stuff.

This follows from the fact that the trapezoid is cyclic.

e) $\triangle ADE$ is congruent to $\triangle BCE$ where E is the intersection of AD, BC.

This follows by many equal lengths and SAS.