In triangle ABC, let P and Q be the midpoints of sides AB and AC, respectively. By the Midpoint Theorem, the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length. Therefore, since PQ connects the midpoints P and Q, it follows that PQ is parallel to side BC of triangle ABC. This establishes that PQ is parallel to BC, as required.
Given ABE, ADC, BD bisescts angle ABC and BD is parallel to EC prove: Triangle EBC is isoceles
To prove triangle ABC is congruent to triangle EDC by the SAS (Side-Angle-Side) Postulate, you need to confirm that two sides and the included angle of triangle ABC are equal to the corresponding two sides and the included angle of triangle EDC. Specifically, you need to know the lengths of sides AB and AC, and the measure of angle A in triangle ABC, as well as the lengths of sides ED and EC, and the measure of angle E in triangle EDC. Once this information is established, you can demonstrate the congruence between the two triangles.
Three parallel vertical lines. A bit like triangle ABC | triangle DEF, except that the lines are closer together.
First draw a triangle with vertices A, B, and C. Let's let C be at the top and B and A the base of the triangle. A' and B' are on the line drawn parallel to the base and through point vertex C. Place A' on the right of C and B' on the left of C. Of course it works for any triangle, but I am trying to make the picture easy so then you can generalize.if ABC is a triangle then
Let the triangle be ABC and its medians by AX and BY and CZ. Therefore, since AC=AB, and OP=BD. Therefore, By Triangle Equiangular Property, Triangle ABC simliar to Triangle XYZ Therefore, Three times the sum of the squares of sides of triangle equal to 4 times of its median.:) Hope it helped.
Given ABE, ADC, BD bisescts angle ABC and BD is parallel to EC prove: Triangle EBC is isoceles
ASA
The circumcircle of a triangle is the circle that passes through the three vertices. Its center is at the circumcenter, which is the point O, at which the perpendicular bisectors of the sides of the triangle are concurrent. Since our triangle ABC is an isosceles triangle, the perpendicular line to the base BC of the triangle passes through the vertex A, so that OA (the part of the bisector perpendicular line to BC) is a radius of the circle O. Since the tangent line at A is perpendicular to the radius OA, and the extension of OA is perpendicular to BC, then the given tangent line must be parallel to BC (because two or more lines are parallel if they are perpendicular to the same line).
Blah blah blah
It is (10, -2).
(-5, 6)
(9, -5)
(6, -4)
BAD = BCD is the answer i just did it
Three parallel vertical lines. A bit like triangle ABC | triangle DEF, except that the lines are closer together.
ABC is an equilateral triangle with side length equal to 50 cm. BH is perpendicular to AC. MN is parallel to AC. Find the area of triangle BMN if the length of MN is equal to 12 cm.
ABC is an equilateral triangle with side length equal to 50 cm. BH is perpendicular to AC. MN is parallel to AC. Find the area of triangle BMN if the length of MN is equal to 12 cm.