Advertisements
Advertisements
प्रश्न
AD and BC are equal perpendiculars to a line segment AB. If AD and BC are on different sides of AB prove that CD bisects AB.
Advertisements
उत्तर
In ΔAOD and ΔBOC,
∠ AOD = ∠ BOC ....(vertically opposite angles)
∠ DAO = ∠ CBO ....(each 90°)
AD = BC ....(given)
∴ ΔAOD ≅ ΔBOC ...(by AAS congruence criterion)
⇒ AO = BO ...(c.p.c.t.)
⇒ O is the mid-point of AB.
Hence, CD bisects AB.
APPEARS IN
संबंधित प्रश्न
Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:
- ΔAPB ≅ ΔAQB
- BP = BQ or B is equidistant from the arms of ∠A.
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that:
- ΔAMC ≅ ΔBMD
- ∠DBC is a right angle.
- ΔDBC ≅ ΔACB
- CM = `1/2` AB
In Fig. 10.92, it is given that AB = CD and AD = BC. Prove that ΔADC ≅ ΔCBA.
Prove that the perimeter of a triangle is greater than the sum of its altitudes.
ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.
A triangle ABC has ∠B = ∠C.
Prove that: The perpendiculars from the mid-point of BC to AB and AC are equal.
In the given figure, AB = DB and Ac = DC.
If ∠ ABD = 58o,
∠ DBC = (2x - 4)o,
∠ ACB = y + 15o and
∠ DCB = 63o ; find the values of x and y.
ABCD is a parallelogram. The sides AB and AD are produced to E and F respectively, such produced to E and F respectively, such that AB = BE and AD = DF.
Prove that: ΔBEC ≅ ΔDCF.
In the following diagram, ABCD is a square and APB is an equilateral triangle.
- Prove that: ΔAPD ≅ ΔBPC
- Find the angles of ΔDPC.
A point O is taken inside a rhombus ABCD such that its distance from the vertices B and D are equal. Show that AOC is a straight line.
