Question

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.

Answer 1

1) Given : AD = BE = CF and angles A = B = C= 90°

To prove : Triangle ABC is equialteral.

Proof : In ∆ABE and ∆ACF

Angle AEB = AFC ( 90° each )

BE = CF ( Given )

Angle A =Angle A ( Common )

Hence ∆ABE≅∆ACFby AAS congruency

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

In ∆AOE and ∆EOC

OE = OE ( Common )

Angle E = Angle E ( 90° each )

AO = OC [AD=FC, their halfs OA = OC ]

Hence ∆AOE≅∆EOC by RHS congruency

AE = EC ( c.p.c.t )

In ∆ABE and ∆BCE

Angle E = Angle E ( 90° each )

BE = BE ( common )

AE = EC ( proved above )

Hence, ∆ABE≅∆BCE by SAS congruency.

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

As AB=AC and AB=BC, so AC = BC.

Hence, AB = BC = CA.

HENCE PROVED.

2)For any quadrilateral ABCD,ABCD, we can easily prove that

AB+BC+CD+DA>AC+BD......(1)(1)AB+BC+CD+DA>AC+BD......

Now, three cases arise. Either,

AC=BDorAC>BDorAC<BDAC=BDorAC>BDorAC<BD.

If AC=BDAC=BD,then the result follows from (1).

If AC>BD⟹AC+BD>2BDAC>BD⟹AC+BD>2BD and then the result follows from (1).

If BD>AC⟹AC+BD>2ACBD>AC⟹AC+BD>2AC and then the result follows from (1).

HENCE PROVED

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