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
Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.
In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.
How had the position of women improved in our country since independence ? Explain with examples.
If two sides of a triangle are 8 cm and 13 cm, then the length of the third side is between a cm and b cm. Find the values of a and b such that a is less than b.
In the following figure ; AC = CD; ∠BAD = 110o and ∠ACB = 74o.
Prove that: BC > CD.
Name the greatest and the smallest sides in the following triangles:
ΔXYZ, ∠X = 76°, ∠Y = 84°.
Name the smallest angle in each of these triangles:
In ΔXYZ, XY = 6.2cm, XY = 6.8cm and YZ = 5cm
ABCD is a quadrilateral in which the diagonals AC and BD intersect at O. Prove that AB + BC + CD + AD < 2(AC + BC).
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PQ > PS
In the given figure, T is a point on the side PR of an equilateral triangle PQR. Show that RT < QT
