1)AD,BE and CF, the altitudes of triangle ABC are equal. prove that triangle ABC is an equilateral triangle.

2) Prove that the sum of all sides of a quadrilateral is greater than the sum of its diagnols.

AD,BE and CF, the altitudes of triangle ABC are equal. prove that triangle ABC is an equilateral triangle

 

We have AD , BE and CF are the altituds of a triangle ABC

Where , AD = BE= CF

Now , let the area of triangle be a.

therefore area of triangle,

 * BC * AD =   * AC * BE =   * AB * CF = a

 We know AD = BE = CF

So,  * BC * AD =   * AC * AD =   * AB * AD

  BC = AC = AB

Hence, ABC is an equilateral triangle

2) Prove that the sum of all sides of a quadrilateral is greater than the sum of its diagonals

   A   B

 

 

  D   C

Draw any quadrilateral and label the vertices A, B, C, and D with vertex A opposite vertex C and vertex B opposite D.
The sides are AB, BC, CD and DA and the diagonals are AC and BD
You are to prove AB + BC + CD + AD > AC + BD
In figure, you formed the triangles, ACD , ABC, ABD and BDC
You know that the sum of two sides of a triangle is always greater than the two sides
In ACD: AD + CD > AC
In ABC: AB + BC > AC
in ABD: AB + AD > BD
in BDC: BC + CD > BD
Adding these 4 inequalities
2AB + 2BC + 2CD + 2AD > 2AC + 2BD
Dividing by 2
AB+BC+CD+AD = AC + BD 

PROVED

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1) consider triangle ABE and triangle ACF

BE = CF (given)

angle A = angle A (common)

angle AEB = angle AFC = 90degrees

therefore by AAS congruency triangle ABE is congruent to triangle ACF

AB = AC (c.p.c.t.)

similarly, triangle DAC is congruent to triangle CBE

AC = BC (c.p.c.t.)

similarly, triangle BCF is congruent to triangle ABD

AB = BC (c.p.c.t.)

now,

AB = BC

AC = BC

AB = AC

this implies that

AB = BC = AC

therefore the triangle is equilateral

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consider a quadrilateral ABCD where AC and BD are the diagonals

AB+BCis greater than AC .......1 (SUM OF TWO SIDES IS GREATER THAN THE THIRD SIDE)

AD+DCis greater thanAC..........2

AB+ADis greater thanBD ..........3

DC+BCis greater thanBD..........4

ADDING 1,2,3,4

AB+BC+AD+DC+AB+AD+DC+BCis greater than AC+AC+BD+BD

2(AB+BC+CD+DA) is greater than2(AC+BD)

AB+BC+CD+DAis greater than AC+BC

HENCE PROVED

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