Advertisements
Advertisements
प्रश्न
In ΔABC, AE is the bisector of ∠BAC. D is a point on AC such that AB = AD. Prove that BE = DE and ∠ABD > ∠C.
Advertisements
उत्तर
In the ΔABE and ΔADE,
AB = AD ....(Given)
∠BAE = ∠DAE ....(AE is the bisector of ∠BAC)
AE = AE ....(Common side)
∴ ΔABE ≅ ΔADE ....(SAS test)
⇒ BE = DE ....(c.p.c.t.c)
In ΔABD,
AB = AD
⇒ ∠ABD = ∠ADB
∠ADB > ∠C ...(Exterior angle property)
⇒ ∠ABD > ∠C.
APPEARS IN
संबंधित प्रश्न
Show that in a right angled triangle, the hypotenuse is the longest side.
In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.
AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see the given figure). Show that ∠A > ∠C and ∠B > ∠D.
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 the following figure, write BC, AC, and CD in ascending order of their lengths.
"Issues of caste discrimination began to be written about in many printed tracts and essays in India in the late nineteenth century." Support the statement with two suitable examples.
Arrange the sides of the following triangles in an ascending order:
ΔDEF, ∠D = 38°, ∠E = 58°.
For any quadrilateral, prove that its perimeter is greater than the sum of its diagonals.
In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that: SN < SR
In ΔPQR, PS ⊥ QR ; prove that: PQ > QS and PQ > PS
