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Suppose the diagonals meet at a point X.
AB is parallel to DC and BD intersects them
Therefore, angle ABD ( = ABX) = BAC (= BAX)
Therefore, in triangle ABX, the angles at the ends of AB are equal => the triangle is isosceles and so AX = BX.
AB is parallel to DC and AC intersects them
Therefore, angle ACD ( = XCD) = BDC (= XDC)
Therefore, in triangle CDX, the angles at the ends of CD are equal => the triangle is isosceles and so CX = DX.
Therefore AX + CX = BX + DX or, AC = BD.
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Show that if diagonals of a quadrilateral bisects each other then it is a rhombus?
This cannot be proven, because it is not generally true. If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. And conversely, the diagonals of any parallelogram bisect each other. However not every parallelogram is a rhombus.However, if the diagonals are perpendicular bisectors, then we have a rhombus.Consider quadrilateral ABCD, with diagonals intersecting at X, whereAC and BD are perpendicular;AX=XC;BX=XD.Then angles AXB, BXC, CXD, DXA are all right angles and are congruent.By the ASA theorem, triangles AXB, BXC, CXD and DXA are all congruent.This means that AB=BC=CD=DA.Since the sides of the quadrilateral ABCD are congruent, it is a rhombus.
How do you prove that the diagonals and either base of an isosceles trapezoid form an isosceles triangle?
Consider the isosceles trapezium ABCD (going clockwise from top left) with AB parallel to CD. And let the diagonals intersect at O Since it is isosceles, AD = BC and <ADC = <BCD (the angles at the base BC). Now consider triangles ADC and BCD. AD = BC The side BC is common and the included angles are equal. So the two triangles are congruent. and therefore <ACD = <BDC Then, in triangle ODC, <OCD (=<ACD = <BDC) = <ODC ie ODC is an isosceles triangle. The triangle formed at the other base can be proven similarly, or by the fact that, because AB CD and the diagonals act as transversals, you have equal alternate angles.
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.
Prove diagonals are equal in a rectangle?
Suppose ABCD is a rectangle.Consider the two triangles ABC and ABDAB = DC (opposite sides of a rectangle)BC is common to both trianglesand angle ABC = 90 deg = angle DCBTherefore, by SAS, the two triangles are congruent and so AC = BD.
Write a two column proof for each of these problems given line ab is congruent to line ad and line ca is congruent to line ea prove angle abc is congruent to angle ade?
There cannot be a proof since the statement need not be true.
