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 the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
In the given figure, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR >∠PSQ.
ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
Complete the hexagonal and star shaped rangolies (see the given figures) by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?
In ΔABC, BC produced to D, such that, AC = CD; ∠BAD = 125° and ∠ACD = 105°. Show that BC > CD.
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PR > PS
In the given figure, T is a point on the side PR of an equilateral triangle PQR. Show that PT < QT
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.
