Advertisements
Advertisements
प्रश्न
In the following figure, BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC. Show that ∆ABC ≅ ∆DEF.
उत्तर
Given: In the following figure, BA ⊥ AC, DE ⊥ DF such that BA = DE and BF = EC.
To show: ∆ABC ≅ ∆DEF
Proof: Since, BF = EC
On adding CF both sides, we get
BF + CF = EC + CF
⇒ BC = EF ...(i)
In ∆ABC and ∆DEF,
∠A = ∠D = 90° ...[∵ BA ⊥ AC and DE ⊥ DF]
BC = EF ...[From equation (i)]
And BA = DE ...[Given]
∴ ∆ABC ≅ ∆DEF ...[By RHS congruence rule]
APPEARS IN
संबंधित प्रश्न
ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show that
- ΔABD ≅ ΔACD
- ΔABP ≅ ΔACP
- AP bisects ∠A as well as ∠D.
- AP is the perpendicular bisector of BC.
AD is an altitude of an isosceles triangles ABC in which AB = AC. Show that
- AD bisects BC
- AD bisects ∠A
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see the given figure). Show that:
- ΔABM ≅ ΔPQN
- ΔABC ≅ ΔPQR
In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.
Prove that in a quadrilateral the sum of all the sides is greater than the sum of its diagonals.
ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC, AB = AC and DB = DC. Show that AD is the perpendicular bisector of BC.
Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.
In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.
Two lines l and m intersect at the point O and P is a point on a line n passing through the point O such that P is equidistant from l and m. Prove that n is the bisector of the angle formed by l and m.
ABCD is a quadrilateral such that diagonal AC bisects the angles A and C. Prove that AB = AD and CB = CD.
