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If bisects the angle ACD then B is the midpoint of AD?
That is not necessarily true.
If A and B are two points in the plane the perpendicular bisector of line segment AB is the set of all points equidistant from A and B?
Let us consider a line l such that it is the perpendicular bisector of line segment AB and the line intersects at point C.Let us take any point on line l(say, D). Join A to D and B to D.Now we have two triangles ACD and BCD.Now, in triangles ACD & BCD, we haveCD = CD (Common side)�ACD = �BCD (Right angle)AC = BC (Since l bisects AB)According to Side-Angle-Side criteria: Both triangles are congruent.Since both triangles are congruent, therefore AD = BD.So, l is the set of all points equidistant from A & B.Hence proved.
Angle bisector of angleA of triangleABC is perpendicular to BC prove it is isosceles?
Let D represent the point on BC where the bisector of A intersects BC. Because AD bisects angle A, angle BAD is congruent to CAD. Because AD is perpendicular to BC, angle ADB is congruent to ADC (both are right angles). The line segment is congruent to itself. By angle-side-angle (ASA), we know that triangle ADB is congruent to triangle ADC. Therefore line segment AB is congruent to AC, so triangle ABC is isosceles.
If the bisector of the vertical angle of the triangle bisects the base prove that the triangle is isosceles?
show that the bisector of the vertical angle of an isosceles triangle bisects the base
If segment AD perpendicular segment CE m angle 1 m angle 6 and m angle 1 37 find m angle GMD?
143
If bisects the angle ACD then B is the midpoint of AD?
That is not necessarily true.
What is the proof that equiangular triangle is also called equilateral triangle?
Label the triangle ABC. Draw the bisector of angle A to meet BC at D. Then in triangles ABD and ACD, angle ABD = angle ACD (equiangular triangle) angle BAD = angle CAD (AD is angle bisector) so angle ADB = angle ACD (third angle of triangles). Also AD is common. So, by ASA, triangle ABD is congruent to triangle ACD and therefore AB = AC. By drawing the bisector of angle B, it can be shown that AB = BC. Therefore, AB = BC = AC ie the triangle is equilateral.
What bisects an angle?
The Angle Bisector Theorem states that given triangle and angle bisector AD, where D is on side BC, then . Likewise, the converse is also true. Not sure if this is what you want?
Argument to showing that an equilateral triangle cannot have a right angle?
If you have an equilateral triangle ABC, then draw the line from A to D, the mid point of BC. Then in trangles ABD and ACD, AB = AC (equilateral), BD = DC (D is midpoint), and AD is common so these two triangles are congruent and so angle ABD = angle ACD. That is, angle ABC = angle ACB. Similarly the third angle can be shown to be the same. Thus an equilateral triangle is also equiangular. Now, sum of the interior angles of any triangle is 180 degrees. So since sum of three equal angles measures is 180 degrees, they must be 60 degrees each, i.e. NOT 90 degrees so there is no right angle..
How do i prove if the base angles of a triangle are congruent then the triangle is isosceles?
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.
In a right angle triangle angle b is 90 d is mid point of side ac ac10 find out ad and how?
If AC = 10 units and D is the midpoint of AC then AD = AC/2 = 5 units!
If A and B are two points in the plane the perpendicular bisector of line segment AB is the set of all points equidistant from A and B?
Let us consider a line l such that it is the perpendicular bisector of line segment AB and the line intersects at point C.Let us take any point on line l(say, D). Join A to D and B to D.Now we have two triangles ACD and BCD.Now, in triangles ACD & BCD, we haveCD = CD (Common side)�ACD = �BCD (Right angle)AC = BC (Since l bisects AB)According to Side-Angle-Side criteria: Both triangles are congruent.Since both triangles are congruent, therefore AD = BD.So, l is the set of all points equidistant from A & B.Hence proved.
What is the coordinate of the midpoint of segment AD?
-2.5
If the median to a side of a triangle is also an altitude to that side then the triangle is isosceles How do you write this Proof?
Let the triangle be ABC and suppose the median AD is also an altitude.AD is a median, therefore BD = CDAD is an altitude, therefore angle ADB = angle ADC = 90 degreesThen, in triangles ABD and ACD,AD is common,angle ADB = angle ADCand BD = CDTherefore the two triangles are congruent (SAS).And therefore AB = AC, that is, the triangle is isosceles.
Angle bisector of angleA of triangleABC is perpendicular to BC prove it is isosceles?
Let D represent the point on BC where the bisector of A intersects BC. Because AD bisects angle A, angle BAD is congruent to CAD. Because AD is perpendicular to BC, angle ADB is congruent to ADC (both are right angles). The line segment is congruent to itself. By angle-side-angle (ASA), we know that triangle ADB is congruent to triangle ADC. Therefore line segment AB is congruent to AC, so triangle ABC is isosceles.
Can the diffference of two sides of a triangle be equal to less than a third side?
The difference between two sides of a triangle will always be less than the third side.Let ABC be a triangle where AC > AB, extend side AB to point D so that AD = AB + (AC-AB) = AC. Therefore, since AC = AD, triangle ADC is a isosceles where angles ADC and ACD are equal.In the triangle BCD, angle BCD < BDC, Since, angles BCD is part of angle ACD (ACB + BCD ) and angle ACD is equal to BDC .Therefore, using the knowledge that, in a triangle, side opposite a greater angle is always greater than the side opposite a smaller angle, it is proved that, the difference between two sides is always lesser than the third side.In an isosceles, the difference of two sides is zero, since the sides are equal. The third side would always be greater than zero to form a triangle. The same logic can be applied to an equilateral triangle.
In a gm ABCD the bisector of angle A also bisect BC at X. Prove tht AD2 AB?
AX bisects angle DAB so angles DAX and XAB are equal. .. .. .. .. .. .. .. .. (i)DA is parallel to CB and AX is an intercept.So angle DAX and AXB are alternate angles and therefore angles DAX and AXB are equal.Therefore, by (i) angles XAB and AXB are equal.Thus triangle BAX is isoscelestherefore AB = BX.BX = 1/2*BC = 1/2*AD (since ABCD is a parallelogram).Therefore AB = BX = AD/2.
