HELPPP PLEASEEE !!! John draws AABD, with CD as the perpendicular bisector of AB such that AACD and ABCD are congruent to each o
ther.
DO
Which statement can John use this figure to prove?
Every isosceles triangle is a right triangle.
The base angles of any triangle are congruent.
The points on the perpendicular bisector of a side of a triangle are equidistant from all of the vertices of the triangle.
The points on the perpendicular bisector of a side of a triangle are equidistant from the vertices of the side it bisects.
DO
Which statement can John use this figure to prove?
Every isosceles triangle is a right triangle.
The base angles of any triangle are congruent.
The points on the perpendicular bisector of a side of a triangle are equidistant from all of the vertices of the triangle.
The points on the perpendicular bisector of a side of a triangle are equidistant from the vertices of the side it bisects.
1 answer:
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• Angles DXC and AXB form a vertical pair, so they are congruent and have the same measure.
• ∆ABD is isosceles, since it's given that AD and BD are congruent. This means the "base angles" BAD and ABD have the same measure; call this measure <em>x</em>.
• The measure of angle ADB can be computed by using the inscribed angle theorem, which says
m∠ADB = 1/2 (100°) = 50°
(that is, it's half the measure of the subtended arc AB whose measure is 100°)
• The interior angle to any triangle sum to 180° in measure. So we have in ∆ABD,
m∠ADB + 2<em>x</em> = 180°
Solve for <em>x</em> :
50° + 2<em>x</em> = 180°
2<em>x</em> = 130°
<em>x</em> = 65°
• Use the inscribed angle theorem again to find the measure of angle BAC. This will be half the measure of the subtended arc BC, so
m∠BAC = 1/2 (50°) = 25°
• Now in ∆ABX, we have
m∠AXB + 25° + 65° = 180°
m∠AXB = 90°
Hence m∠DXC = 90°.
