Advertisements
Advertisements
प्रश्न
ΔABC in a isosceles triangle with AB = AC. D is a point on BC produced. ED intersects AB at E and AC at F. Prove that AF > AE.
Advertisements
उत्तर
∠AEF > ∠ABC ...(Exterior angle property)
∠AEF = ∠DFC
∠ACB > ∠DFC ...(Exterior angle property)
⇒ ∠ACB > ∠AFE
Since AB = AC
⇒ ∠ACB = ∠ABC
So, ∠ABC > ∠AFE
⇒ ∠AEF > ∠ABC > ∠AFE
that is ∠AEF > ∠AFE
⇒ AF > AE.
APPEARS IN
संबंधित प्रश्न
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
In a huge park people are concentrated at three points (see the given figure):
A: where there are different slides and swings for children,
B: near which a man-made lake is situated,
C: which is near to a large parking and exit.
Where should an ice-cream parlour be set up so that maximum number of persons can approach it?
(Hint: The parlor should be equidistant from A, B and C)
In the following figure, write BC, AC, and CD in ascending order of their lengths.
“Caste inequalities are still prevalent in India.” Examine the statement.
Prove that the hypotenuse is the longest side in a right-angled triangle.
D is a point on the side of the BC of ΔABC. Prove that the perimeter of ΔABC is greater than twice of AD.
In ABC, P, Q and R are points on AB, BC and AC respectively. Prove that AB + BC + AC > PQ + QR + PR.
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PQ > PS
Prove that in an isosceles triangle any of its equal sides is greater than the straight line joining the vertex to any point on the base of the triangle.
In ΔABC, D is a point in the interior of the triangle. Prove that DB + DC < AB + AC.
