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There cannot be a proof since the statement need not be true.
Suppose you have triangle ABC with base BC, and angle B = angle C. Draw the altitude AD.Considers triangles ABD and ACDangle ABD = angle ACD (given)angle ADB = 90 deg = angle ACDtherefore angle BAD = angle CADAlso the side AD is common to the two triangles.Therefore triangle ABD is congruent to triangle ACD (ASA) and so AB = AC.That is, triangle ABC is isosceles.
Let's call the two angles angle 1 and angle 2. We are given that angle 1 and angle 4 form a linear angle and that angle 2 and angle 4 form a linear angle. Because linear angles measure 180 degrees, we arrive at: m<1 + m<4 = 180 m<2 + m<4 = 180. By subtracting the second equation from the first, we get: m<1 - m<2 = 0. And finally: m<1 = m<2. Thus, angle 1 is congruent to angle 2.
The set of all points in a plane that are equidistant from the two sides of a given angle
Yeah or yeha
Given: AD perpendicular to BC; angle BAD congruent to CAD Prove: ABC is isosceles Plan: Principle a.s.a Proof: 1. angle BAD congruent to angle CAD (given) 2. Since AD is perpendicular to BC, then the angle BDA is congruent to the angle CDA (all right angles are congruent). 3. AD is congruent to AD (reflexive property) 4. triangle BAD congruent to triangle CAD (principle a.s.a) 5. AB is congruent to AC (corresponding parts of congruent triangles are congruent) 6. triangle ABC is isosceles (it has two congruent sides)
Without a visual or more information, I'm guessing that the picture is of angles 1 and 2 that are consecutive (share an angle side) and a separate picture of consecutive angles 3 and 4. With that said: 1) angle 2 congruent to angle 3................1) given 2) angle 1 is supplementary to angle 2....2) If angles are next to each other --> supps angle 3 is supplementary to angle 4 3) angle 1 congruent angle 4..............3) If supps to congruents angles ---> congruent
There cannot be a proof since the statement need not be true.
Angle cpb is given as 17 degrees, and it's inside angle apb. Additionally, angle cpb is congruent to angle apc. That means angle apb is twice angle cpb, or twice 17 degrees, or 34 degrees.
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Complementary angles are those that add up to 90° Thus all angles complementary to a given angle must all be the same angle (90° - the_given_angle), ie they are all congruent angles.
Since a bisector splits an angle into two congruent angles, it goes directly down the center. Therefore there can only be one bisector for any given angle.
True. Only if the given angle is between the two sides will the two triangles guarantee to be congruent (SAS), unless the given angle is a right angle (90°) in which case you now have RHS (Right-angle, Hypotenuse, Side) which does guarantee congruence.
The opposite angles of a rhombus are congruent. So the angle opposite to the given angle is also 35 degrees. The consecutive angles of a rhombus are supplementary (add up to 180 degrees). So the supplement angle of the given angle is 145 degrees (180 - 35), and the angle opposite to that angle also will be 145 degrees.
To find the complementary angle, you subtract 90 by the first given complement angle. To find the supplementary angle, you subtract 180 by the first given supplement angle.
Given two intersecting lines, the two angles opposite each other have the same measure and are congruent.
Suppose you have triangle ABC with base BC, and angle B = angle C. Draw the altitude AD.Considers triangles ABD and ACDangle ABD = angle ACD (given)angle ADB = 90 deg = angle ACDtherefore angle BAD = angle CADAlso the side AD is common to the two triangles.Therefore triangle ABD is congruent to triangle ACD (ASA) and so AB = AC.That is, triangle ABC is isosceles.